AP Inter 1st Year

Referring to the AP Inter 1st Year Maths Study Material Chapter 12 Introduction to 3D Geometry Exercise 12c Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Introduction to 3D Geometry Solutions Exercise 12c

I. Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

Question 1.
(x + a)
Solution:
Let f(x) = (x + a). Accordingly, f(x + h) = x + h + a
By first principle,
f'(x) = \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
= \(\lim _{h \rightarrow 0} \frac{x+h+a-x-a}{h}\)
= \(\lim _{h \rightarrow 0}\left(\frac{h}{h}\right)\)
= \(\lim _{h \rightarrow 0}\) (1)
= 1

Question 2.
(px + q)(\(\frac{r}{x}\) + s)
Solution:
f(x) = (px + q)(\(\frac{r}{x}\) + s), By Leibnitz product rule,
f'(x) = (px + q)(\(\frac{r}{x}\) + s)’ + (\(\frac{r}{x}\) + s)(px + q)’
= (px + q)(rx^-1 + s)’ + (\(\frac{r}{x}\) + s)(p)
= (px + q)(-rx^-2) + (\(\frac{r}{x}\) + s)p
= \(\frac{-\mathrm{pr}}{\mathrm{x}}-\frac{\mathrm{qr}}{\mathrm{x}^2}+\frac{\mathrm{pr}}{\mathrm{x}}\) + ps
= ps – \(\frac{q r}{x^2}\)

Question 3.
(ax + b)(cx + d)²
Solution:
Let f(x) = (ax + b)(cx + d)²
By Leibnitz Product rule f'(x) = (ax + b)\(\frac{d}{d x}\)(cx + d)² + (cx + d)²\(\frac{d}{d x}\)(ax + d)
= (ax + b)\(\frac{d}{d x}\)(c²x² + 2cdx + d²) + (cx + d)²\(\frac{d}{d x}\)(ax + b)
= (ax + b)[\(\frac{d}{d x}\)(c²x²) + \(\frac{d}{d x}\)(2cdx) + \(\frac{d}{d x}\)(d²)] + (cx + d)²[\(\frac{d}{d x}\)(ax) + \(\frac{d}{d x}\)(b)]
= (ax + b)(2c²x + 2cd) + (cx + d)²a
= 2c(ax + b)(cx + d) + a(cx + d)²

Question 4.
\(\frac{a x+b}{c x+d}\)
Solution:
Let f(x) = \(\frac{a x+b}{c x+d}\), By quotient rule,
f'(x) = \(\frac{(c x+d) \frac{d}{d x}(a x+b)-(a x+b) \frac{d}{d x}(c x+d)}{(c x+d)^2}=\frac{(c x+d)(a)-(a x+b)(c)}{(c x+d)^2}\)

= \(\frac{a c x+a d-a c x-b c}{(c x+d)^2}=\frac{a d-b c}{(c x+d)^2}\)

Question 5.
\(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\)
Solution:
Let f(x) = \(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}=\frac{\frac{x+1}{x}}{\frac{x-1}{x}}=\frac{x+1}{x-1}\), where x ≠ 0

By quotient rule, f(x) = \(\frac{(x-1) \frac{d}{d x}(x+1)-(x+1) \frac{d}{d x}(x-1)}{(x-1)^2}\), x ≠ 0, 1
= \(\frac{(x-1)(1)-(x+1)(1)}{(x-1)^2}\), x ≠ 0, 1
= \(\frac{x-1-x-1}{(x-1)^2}\), x ≠ 0, 1
= \(\frac{-2}{(x-1)^2}\), x ≠ 0, 1

Question 6.
\(\frac{1}{a x^2+b x+c}\)
Solution:
Let f(x) = \(\frac{1}{a x^2+b x+c}\), By quotient rule,
f'(x) = \(\frac{\left(a x^2+b x+c\right) \frac{d}{d x}(1)-\frac{d}{d x}\left(a x^2+b x+c\right)}{\left(a x^2+b x+c\right)^2}\)
= \(\frac{\left(a x^2+b x+c\right)(0)-(2 a x+b)}{\left(a x^2+b x+c\right)^2}=\frac{-(2 a x+b)}{\left(a x^2+b x+c\right)^2}\)

Question 7.
\(\frac{a x+b}{p x^2+q x+r}\)
Solution:
Let f(x) = \(\frac{a x+b}{p x^2+q x+r}\), By quotient rule,

Question 8.
\(\frac{p x^2+q x+r}{a x+b}\)
Solution:
Let f(x) = \(\frac{p x^2+q x+r}{a x+b}\), By quotient rule,

Question 9.
\(\frac{a}{x^4}-\frac{b}{x^2}\) + cos x
Solution:
Let f(x) = \(\frac{a}{x^4}-\frac{b}{x^2}\) + cos x

f'(x) = \(\frac{d}{d x}\left(\frac{a}{x^4}\right)-\frac{d}{d x}\left(\frac{b}{x^2}\right)+\frac{d}{d x}\)(cos x)
= a\(\frac{\mathrm{d}}{\mathrm{dx}}\)(x^-4) – b\(\frac{\mathrm{d}}{\mathrm{dx}}\)(x^-2) + \(\frac{\mathrm{d}}{\mathrm{dx}}\)(cos x)

= a(-4x^-5) – b(-2x^-3) + (-sin x) [∵ \(\frac{d}{d x}\)(xⁿ) = nx^n-1 and \(\frac{d}{d x}\)(cos x) = -sin x]

= \(\frac{-4 a}{x^5}+\frac{2 b}{x^3}\) – sin x

Question 10.
4√x – 2
Solution:
Let f(x) = 4√x – 2
f'(x) = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(4√x – 2) = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(4√x) – \(\frac{\mathrm{d}}{\mathrm{dx}}\)(2)
= 4\(\frac{1}{2}\)(x^{\(\frac{1}{2}\)}) – 0
= 4(\(\frac{1}{2}\)x^{\(\frac{1}{2}\) – 1})
= (2x\(\frac{-1}{2}\))
= \(\frac{2}{\sqrt{x}}\)

Question 11.
(ax + b)ⁿ
Solution:
Method I : Let y = (ax + b)ⁿ
Differentiating w.r.t ‘x’
\(\frac{d y}{d x}=\frac{d}{d x}\) (ax + b)ⁿ
⇒ \(\frac{d y}{d x}\) = n(ax + b)^n-1 [/latex]\frac{d}{d x}[/latex](ax + b)
⇒ \(\frac{d y}{d x}\) = an(ax + b)^n-1

Method II : Let f(x) = (ax + b)ⁿ
Accordingly, f(x + h) = {a(x + h) + b}ⁿ = (ax + ah + b)ⁿ
By first principle,
f'(x) = \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{(a x+a h+b)^n-(a x+b)^n}{h}\)

Question 12.
(ax + b)ⁿ (cx + d)^m
Solution:
Method I : Let y = (ax + b)ⁿ(cx + d)^m
Differentiate w.r.t ‘x’
\(\frac{d y}{d x}\)(ax + b)ⁿ m(cx + d)^m-1 c + (cx + d)ⁿ n (ax + b)^n-1a
\(\frac{d y}{d x}\) = (ax + b)^n-1 (cx + d)^m-1 [mc(ax + b) + na (cx + d)]

Methqd II: Let f(x) = (ax + b)ⁿ (cx + d)^m
By Leibnitz product rule,
f'(x) = (ax + b)ⁿ \(\frac{d}{d x}\) (cx + d)^m + (cx + d)^m. \(\frac{d}{d x}\) (ax + b)ⁿ ……….(1)
Now, Let f₁(x) = (cx + d)^m
Accordingly, f(x + h) = (cx + ch + d)^m

Hence, \(\frac{d}{d x}\)(cx + d)^m = mc(cx + d)^m-1 ………(2)
Similarly, \(\frac{d}{d x}\)(ax + b)ⁿ = na(Ax + b)^n-1 ………….(3)
Therefore, from (1), (2) and (3), we obtain
f'(x) = (ax + b)ⁿ{mc(cx + d)^m-1} + (cx + b)^m{na(ax + b)^n-1}
= (ax + b)^n-1{(cx + d)^m-1} + (cx + b)^m{na(ax + b)^n-1}
= (ax + b)^n-1(cx + d)^m-1[mc(ax + b) + na(cx + d)]

Question 13.
sin(x + a)
Solution:
Let f(x) = sin(x + a). Accordingly, f(x + h) = sin(x + h + a)
By first principle,

Question 14.
Find the derivatives of the following functions.
i) cotⁿ x
Solution:
cotⁿx
Let y = cotⁿ x.
Differentiate w.r.t ‘x’
\(\frac{d y}{d x}\) = -n cot^n-1x. cosec²x

ii) cosec⁴ x
Solution:
Let y = cosec⁴ x.
Differentiate w.r.t ‘x’
\(\frac{d y}{d x}\) = 4 cosec³x\(\frac{1}{2}\)(cosec x)
⇒ \(\frac{d y}{d x}\) = -4 cosec⁴ x cot x

iii) sin^mx . cosⁿx
Solution:
Let y = sin^mx . cosⁿx
Differentiate w.r.t ‘x’
\(\frac{d y}{d x}=\frac{d}{d x}\){sin^mx . cosⁿx} = sin^mx \(\frac{d}{d x}\)cosⁿx + cosⁿx\(\frac{d}{d x}\)sin^mx
\(\frac{d y}{d x}\) = msin^m-1x cosⁿ⁺¹x – ncos^n-1x sin^m+1x

iv) sin mx. cos nx
Solution:
Let y = sin mx. cos nx
Differentiate w.r.t ‘x’
\(\frac{d y}{d x}\) = sin mx. \(\frac{d}{d x}\)cos nx + cos nx \(\frac{d}{d x}\)sin mx

\(\frac{d y}{d x}\) = mcosmx cosnx – nsinmx sinnx

v) log (tan 5x)
Solution:
Let y = log (tan 5x)
Differentiate w.r.t ‘x’
⇒ \(\frac{d y}{d x}=\frac{1}{\tan 5 x}\) × sec²5x × 5
⇒ \(\frac{d y}{d x}=\frac{5 \sec ^2 5 x}{\tan 5 x}\)
⇒ \(\frac{d y}{d x}=\frac{5}{\sin 5 x \cos 5 x}\)
⇒ \(\frac{d y}{d x}\) = 10 cosec 10x

vi) log \(\left(\frac{x^2+x+2}{x^2-x+2}\right)\)
Solution:
Let y = log \(\left(\frac{x^2+x+2}{x^2-x+2}\right)\)
⇒ y = log(x² + x + 2) – log(x² – x + 2)
Differentiate w.r.t ‘x’

vii) cos (log x + e^x)
Solution:
Let y = cos (log x + e^x)
Differentiate w.r.t ‘x’
\(\frac{d y}{d x}\) = -sin(log x + e^x)[\(\frac{1}{x}\) + e^x]

II.

Question 1.
Find the derivative of the following function. from first principle.
i) -x
Solution:
Let f(x) = -x
Accordingly, f(x + h) = -(x + h)
By first principle, f'(x) = \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
= \(\lim _{h \rightarrow 0} \frac{-(x+h)+(x)}{h}=\lim _{h \rightarrow 0} \frac{-x-h+x}{h}\)
= \(\lim _{h \rightarrow 0} \frac{-h}{h}=\lim _{h \rightarrow 0}(-1)\)
= -1

ii) (-x)^-1
Solution:
Let f(x) = (-x)^-1 = \(\frac{-1}{x}\)
Accordingly, f(x + h) = \(\frac{-1}{(x+h)}\)
By first principle, f'(x) = \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
= \(\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{-1}{(x+h)}-\left(\frac{-1}{x}\right)\right]=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{-x+(x+h)}{x(x+h)}\right]\)
= \(\lim _{h \rightarrow 0} \frac{1}{x(x+h)}=\frac{1}{x(x)}=\frac{1}{x^2}\)

iii) sin(x + 1)
Solution:
Let f(x) = sin(x + 1)
Accordingly, f(x + h) = sin(x + h + 1)

iv) cos(x – \(\frac{\pi}{8}\))
Solution:
Let f(x) = cos(x – \(\frac{\pi}{8}\))
Accordingly, f(x + h) = cos(x + h – \(\frac{\pi}{8}\))
By first principle, f'(x) = \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q. r and s are fixed non-zero constants and m and n are integers).

Question 2.
cosec x cot x
Solution:
Let y = cosec x cot x.
Differentiate w.r.t ‘x’
\(\frac{d y}{d x}\) = cosec x \(\frac{\mathrm{d}}{\mathrm{dx}}\) cot x + cot x\(\frac{\mathrm{d}}{\mathrm{dx}}\)cosecx
= cosec x(-cosec²x) + cot x(- cosecx cot x)
⇒ -cosec³x – cot²x cosec x.

Question 3.
\(\frac{\cos x}{1+\sin x}\)
Sol.
Let f(x) = \(\frac{\cos x}{1+\sin x}\)
By quotient rule,

Question 4.
\(\frac{\sin x+\cos x}{\sin x-\cos x}\)
Solution:
Let f(x) = \(\frac{\sin x+\cos x}{\sin x-\cos x}\). By quotient rule,

Question 5.
\(\frac{\sec x-1}{\sec x+1}\)
Solution:
Let f(x) = \(\frac{\sec x-1}{\sec x+1}\)

Question 6.
Solution:
Method 1: Let y = sinⁿx
Differentiate w.r.t ‘x’
\(\frac{d y}{d x}=\frac{d}{d x}\) sinⁿx = n sin^n-1x \(\frac{d}{d x}\) (sin x)
\(\frac{d y}{d x}\) = n sin^n-1 x cos x

Method II : Let y = sinⁿx
Accordingly, for n = 1, y = sin x
Therefore, \(\frac{1}{2}\) = cos x, i.e.. \(\frac{1}{2}\)sin x = cos x
For n = 2, y = sin²x; \(\frac{d y}{d x}=\frac{d}{d x}\) (sin x sin x)
= (sin x)’ (sin x)+ sin x(sin x)’ [By Leibnltz product rule]
= cos x sin x + sin x cos x
= 2 sin x cos x ……….. (1)
For n = 3, y sin³x; \(\frac{d y}{d x}=\frac{d}{d x}\)(sin x sin²x)
= (sin x) sin²x + sin x(sin²x) [By Leibnitz product rule]
= cos x sin²x + sin x (2sin x cos x) [using (1)]
= cos x sin²x + 2 sin²x cos x
= 3 sin²x cos x
We assert that \(\frac{\mathrm{d}}{\mathrm{dx}}\)(sinⁿx) = n sin^n-1x cos x
Let our assertion be true for n = k
i.e., \(\frac{\mathrm{d}}{\mathrm{dx}}\)(sin^kx) = k sin^(k-1)x cos x ….. (2)

Consider, \(\frac{d}{d x}\)(sin^k+1x) = \(\frac{d}{d x}\)(sin x sin^(k)x)
= (sin x)’ sin^kx + sin x (sin^kx)’ [By Leibnitz product rule]
= cos x sin^kx + sin x(k sin^k-1 x cos x) [using (2)]
= cos x sin^kx + k sin^kx cos x-1
= (k + 1)sin^kx cos x
Thus, our asssertion is true for n = k + 1
Hence, by mathematicall induction, \(\frac{d}{d x}\)(sinⁿx) = n sin^(n-1)x cos x

Question 7.
\(\frac{a+b \sin x}{c+d \cos x}\)
Solution:
Let f(x) = \(\frac{a+b \sin x}{c+d \cos x}\), By quotient rule,

Question 8.
\(\frac{\sin (x+a)}{\cos x}\)
Solution:
Let f(x) = \(\frac{\sin (x+a)}{\cos x}\), By quotient rule,

Question 9.
x⁴(5 sin x – 3 cos x)
Solution:
Let f (x) = x⁴(5sin x – 3 cosx)
By product rule,
f ‘(x) = x⁴ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (5 sin x – 3 cos x) + (5sin x – 3 cos x) \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x⁴)
= x⁴[5\(\frac{\mathrm{d}}{\mathrm{dx}}\)(sin x) – 3\(\frac{\mathrm{d}}{\mathrm{dx}}\)(cos x)] + (5 sin x – 3 cos x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)(x⁴)
= x⁴ [5 cos x – 3(- sin x)] + (5 sin x – 3 cos x) (4x³)
= x³ [5x cos x + 3x sinx + 20 sinx – 12 cos x]

Question 10.
(x² + 1) cos x
Solution:
Let f(x) = (x² + 1) cosx
By product rule,
f ‘(x) = (x² + 1) \(\frac{\mathrm{d}}{\mathrm{dx}}\) (cos x) + cos x \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x² + 1)
= (x² + 1)(-sinx) + cosx (2x)
= -x² sin x – sin x + 2x cos x

Question 11.
(ax² + sin x) (p + q cos x)
Solution:
Let f(x) = (ax² + sin x) (p + q cos x)
By product rule
f'(x) = (ax² + sin x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)(p + q cos x) + (p + q cos x) \(\frac{\mathrm{d}}{\mathrm{dx}}\) (ax² + sin x)
= (ax² + sinx) (-q sin x) + (p + q cos x) (2ax + cos x)
= – q sin x(ax² + sin x) + (p + q cos x)(2ax + cos x)

Question 12.
(x + cos x) (x – tan x)
Solution:
Let y = (x + cos x) (x – tan x)
Differentiate w.r.t ‘x’
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (x + cos x) \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x – tan x) + (x – tan x) \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x + cos x)

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (x + cos x)(1 – sec²x) + (x – tan x) ( 1 – sin x)

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -tan²x(x + cos x) + (x – tan x)(1 – sin x)

Question 13.
\(\frac{4 x+5 \sin x}{3 x+7 \cos x}\)
Solution:
Let f(x) = \(\frac{4 x+5 \sin x}{3 x+7 \cos x}\)
By product rule

Question 14.
\(\frac{x^2 \cos \left(\frac{\pi}{4}\right)}{\sin x}\)
Solution:
Let f(x) = \(\frac{x^2 \cos \left(\frac{\pi}{4}\right)}{\sin x}\), By wuotient rule,

Question 15.
\(\frac{x}{1+\tan x}\)
Solution:
Let f(x) = \(\frac{x}{1+\tan x}\)
f'(x) = \(\frac{(1+\tan x) \frac{d}{d x}(x)-(x) \frac{d}{d x}(1+\tan x)}{(1+\tan x)^2}\)

= \(=\frac{(1+\tan x)-x \sec ^2 x}{(1+\tan x)^2}\)

⇒ f'(x) = \(\frac{1+\tan x-x \sec ^2 x}{(1+\tan x)^2}\)

Question 16.
(x + sec x) (x – tan x)
Solution:
Let f(x) = (x + sec x)(x – tan x). By product rule,
f’(x) = (x+sec x)(x – tan x) + (x – tan x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)(x + sec x)
= (x + sec x) [\(\frac{\mathrm{d}}{\mathrm{dx}}\)(x) – \(\frac{\mathrm{d}}{\mathrm{dx}}\)tan x] + (x – tan x)[\(\frac{\mathrm{d}}{\mathrm{dx}}\)(x) + \(\frac{\mathrm{d}}{\mathrm{dx}}\)sec x]
f’(x) = (x+ sec x) (1 – sec 2x) + (x – tan x) 1 + (sec x tan x)

Question 17.
\(\frac{x}{\sin ^n x}\)
Solution:
Let f(x) = \(\frac{x}{\sin ^n x}\). By quotient rule,
f'(x) = \(\frac{\sin ^n x \frac{d}{d x} x-x \frac{d}{d x} \sin ^n x}{\sin ^{2 n} x}\)
It can be easily shown that \(\frac{d}{d x}\) sinⁿx = n sin^n-1x cos x

Referring to the AP Inter 1st Year Maths Study Material Chapter 12 Introduction to 3D Geometry Exercise 12d Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Introduction to 3D Geometry Solutions Exercise 12d

Differentiate the functions with respect to x in Exercises 1 to 8.

Question 1.
sin(x² + 5)
Solution:
Lety = sin (x² + 5).
Differentiate w.r.t ‘x’
\(\frac{d y}{d x}\) = cos(x + 5) (2x)
⇒ \(\frac{d y}{d x}\) = 2x cos (x² + 5)

Question 2.
cos(sin x)
Solution:
Let y = cos (sin x). Differentiate w.r.t ‘x’
\(\frac{d y}{d x}\) = sin(sin x) cosx

Question 3.
sin(ax + b)
Solution:
Lety = sin(ax + b).
Differentiate w.r.t ‘x’
\(\frac{d y}{d x}\) = cos(ax + b) (ax + b)
⇒ \(\frac{d y}{d x}\) = a cos (ax + b)

Question 4.
sec(tan (√x))
Solution:
Let y = sec (tan (√x)).
Differentiate w.r.t x’
\(\frac{d y}{d x}=\frac{d}{d x}\) sec(tan √x)
⇒ \(\frac{d y}{d x}\) = sec(tan √x) tan(tan √x) \(\frac{d}{d x}\)(tan √x)
= sec(tan √x) tan(tan √x) sec² √x × \(\frac{1}{2 \sqrt{x}}\)
∴ \(\frac{d y}{d x}=\frac{\sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \sec ^2 \sqrt{x}}{2 \sqrt{x}}\)

Question 5.
\(\frac{\sin (a x+b)}{\cos (c x+d)}\)
Solution:
Let y = \(\frac{\sin (a x+b)}{\cos (c x+d)}\)
Differentiate w.r.t ‘x’
\(\frac{d y}{d x}=\frac{\cos (c x+d) \frac{d}{d x} \sin (a x+b)-\sin (a x+b) \frac{d}{d x} \cos (c x+d)}{[\cos (c x+d)]^2}\)

Question 6.
cos x³. sin²(x⁵)
Solution:
Let y = cos x³ sin² (x⁵).
Differentiate w.r.t ‘x’
\(\frac{1}{2}\) = cos x3 \(\frac{1}{2}\) sin2(x5) + sin2(x5) cos x3
since (sin² x⁵) = 2 sin x⁵ sin x⁵ – 2 sin x⁵ cos x⁵ x⁵
= 10 x⁴ sin x⁵ cos x⁵
\(\frac{d}{d x}\) cos x³ = -sin x \(\frac{d}{d x}\)x³ = -3x²sin x³.
\(\frac{d y}{d x}\) = cos³ x(10x sin x⁵ cos x⁵) + sin² x⁵(-3 x² sin x³)
\(\frac{d y}{d x}\) = – 10x⁴ sin x⁵ cos x⁵ cos³ x – 3x² sin x³ sin² x⁵

Question 7.
cos √x
Solution:
Let y = cos √x.
Differentiate w.r.t ‘x’
\(\frac{d y}{d x}\) = -sin √x \(\frac{d}{d x}\)√x
⇒ \(\frac{d y}{d x}=\frac{-\sin \sqrt{x}}{2 \sqrt{x}}\)

Practice AP Inter 1st Year Maths Study Material Chapter 12 Limits and Derivatives MCQ to identify your strengths and weak areas.

AP Inter 1st Year Maths Limits and Derivatives MCQ

Question 1.
\(\frac{1}{2}\)
1) 0
2) -2
3) 2
4) Does not exist
Answer:
4) Does not exist

Explanation:
LHL: \(\lim _{x \rightarrow 1^{-}} \frac{x^2-1}{|x-1|}=\lim _{x \rightarrow 1^{-}} \frac{(x-1)(x+1)}{-(x-1)}=\lim _{x \rightarrow 1}\)-(x + 1) = -2
RHL : \(\lim _{x \rightarrow 1^{+}} \frac{x^2-1}{|x-1|}=\lim _{x \rightarrow 1^{+}} \frac{(x-1)(x+1)}{x-1}=\lim _{x \rightarrow 1}\)(x + 1) = 2
LHL ≠ RHL

Question 2.
\(\lim _{x \rightarrow 0} \frac{1}{x}\)cos<sup-1\(\frac{\left(1-x^2\right)}{1+x^2}\)
1) 0
2) 1
3) 2
4) Does not exist
Answer:
4) Does not exist

Question 3.
\(\lim _{x \rightarrow 0}\)(1 + tan²x)^{\(\frac{1}{2 x^2}\)}
1) e^{\(\frac{1}{2}\)}
2) e
3) e^{\(-\frac{1}{2}\)}
4) 1
Answer:
1) e^{\(\frac{1}{2}\)}

Explanation:
\(\lim _{x \rightarrow 0}\)(1 + tan²x)^{\(\frac{1}{2 x^2}\)} = e^{\(\lim _{x \rightarrow 0} \frac{1}{2 x^2}\left(1+\tan ^2 x-1\right)\)}
= e^{\(\lim _{x \rightarrow 0} \frac{\tan ^2 x}{2 x^2}\)}
= e^{\(\lim _{x \rightarrow 0} \frac{\tan ^2 x}{2 x^2}\)}
= e^{\(\frac{1}{2}\)(1)}
= e^{\(\frac{1}{2}\)} (∵ \(\lim _{x \rightarrow 0}\)f(x)^g(x) = e^{\(\lim _{x \rightarrow 0}(f(x)-1) g(x)\)})

Question 4.
If \(\frac{-\pi}{2}\) < α < \(\frac{\pi}{2}\) then \(\lim _{x \rightarrow 2} \frac{(\cos \alpha)^x+(\sin \alpha)^x-1}{x-2}\)
1) cos²α log cos α
2) sin²α log sin α
3) cos²α log cos α + sin²α log sin α
4) 0
Answer:
3) cos²α log cos α + sin²α log sin α

Explanation:
\(\lim _{x \rightarrow 2} \frac{(\cos \alpha)^x+(\sin \alpha)^x-1}{x-2}\left(\frac{0}{0} \text { form }\right)\)
= \(\lim _{x \rightarrow 2} \frac{(\cos \alpha)^x \log \cos \alpha+(\sin \alpha)^x \log \sin \alpha}{1}\)
= (cos²α)log cos α + sin²α log sin α

Question 5.
If ∆(x) = \(\left|\begin{array}{cc}
e^x & -1 \\
\sin x-1 & 1
\end{array}\right|\) then \(\lim _{x \rightarrow 0} \frac{\Delta(x)}{x}\)
1) 0
2) 1
3) 2
4) -1
Answer:
3) 2

Explanation:
∆(x) = \(\left|\begin{array}{cc}
e^x & -1 \\
\sin x-1 & 1
\end{array}\right|\) = e^x + sin x – 1
\(\lim _{x \rightarrow 0} \frac{\Delta x}{x}=\lim _{x \rightarrow 0} \frac{e^x+\sin x-1}{x}=\lim _{x \rightarrow 0} \frac{e^x-1}{x}+\lim _{x \rightarrow 0} \frac{\sin x}{x}\)
= 1 + 1 = 2

Question 6.
\(\lim _{x \rightarrow 0}\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}\) =
1) \(\frac{1}{\sqrt{a b}}\)
2) \(\sqrt{a b}\)
3) \(\frac{1}{a b}\)
4) ab
Answer:
2) \(\sqrt{a b}\)

Explanation:
\(\lim _{x \rightarrow 0}\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}\) (1^∞ form)
= \(e^{\lim _{x \rightarrow 0}\left(\frac{a^x+b^x}{2}-1\right) \frac{1}{x}}=e^{\lim _{x \rightarrow 0}\left(\frac{a^x+b^x-2}{2 x}\right)}\)
= \(e^{\frac{1}{2}} \lim _{x \rightarrow 0} \frac{a^x-1}{x}+\frac{b^x-1}{x}\)
= e^{\(\frac{1}{2}\)(log a + log b)}
= e^{\(\frac{1}{2}\)log ab}
= e^{log\(\sqrt{a b}\)}
= \(\sqrt{a b}\)

Question 7.
\(\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^3}\) =
1) –\(\frac{1}{2}\)
2) 0
3) 2
4) \(\frac{1}{2}\)
Answer:
4) \(\frac{1}{2}\)

Explanation:
\(\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^3}=\lim _{x \rightarrow 0} \frac{\tan x}{x} \frac{(1-\cos x)}{x^2}=\lim _{x \rightarrow 0} \frac{\tan x}{x} \cdot \lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}\)
= 1 × \(\frac{1}{2}\) = \(\frac{1}{2}\)

Question 8.
If a, b,d > 0 and ≠ 1 then \(\lim _{x \rightarrow 0} \frac{a^x-b^x}{d^x-1}\)
1) \(\log _d\left(\frac{a}{b}\right)\)
2) \(\log _d\left(\frac{b}{a}\right)\)
3) \(\log _{\left(\frac{\mathrm{a}}{\mathrm{~b}}\right)} \mathrm{d}\)
4) \(\log _{\frac{a}{b}} d\)
Answer:
1) \(\log _d\left(\frac{a}{b}\right)\)

Explanation:
\(\lim _{x \rightarrow 0} \frac{a^x-b^x}{d^x-1}=\lim _{x \rightarrow 0} \frac{\frac{a^x-1}{x}-\frac{b^x-1}{x}}{\frac{d^x-1}{x}}\)
= \(\frac{\log a-\log b}{\log d}=\frac{\log \frac{a}{b}}{\log d}\)
= \(\log _d\left(\frac{a}{b}\right)\)

Question 9.
\(\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\frac{1}{\sin x}}\)
1) 1
2) e^-1
3) e^-2
4) e
Answer:
1) 1

Explanation:

Question 10.
\(\frac{1}{2}\)
1) -1
2) 0
3) 1
4) \(\frac{1}{2}\)
Answer:
3) 1

Explanation:

Question 11.
Assertion (A): \(\lim _{x \rightarrow 0} \frac{\sin x}{x}\) = 1
Reason (R) : If \(\lim _{x \rightarrow a}\) f(x) = l, \(\lim _{x \rightarrow a}\)g(x) = m then \(\lim _{x \rightarrow a}\) f(x)g(x) = lm
1) Both A and R are true and R is correct explanation of A.
2) Both A and R are true and R is not correct explanation of A.
3) A is true but R is false.
4) A is false but R is true.
Answer:
2) Both A and R are true and R is not correct explanation of A.

Question 12.
Assertion (A): \(\lim _{x \rightarrow 0} \frac{|x|}{x}\) = 1
Reason (R) : f: R – {0} → R defined by f(x) = \(\frac{|x|}{x}\) has range {-1, 1}
1) Both A and R are true and R is correct explanation of A.
2) Both A and R are true and R is not correct explanation of A.
3) A is true but R is false.
4) A is false but R is true.
Answer:
4) A is false but R is true.

Explanation:
\(\lim _{x \rightarrow 0} \frac{|x|}{x}\) does not exist.

Question 13.
Let f(x) = x + [x], x ∈ R then f’\(\left(\frac{3}{2}\right)\) is equal to
1) 3/2
2) 1
3) 0
4) 5/2
Answer:
2) 1

Explanation:
f(x) – x + 1
⇒ f(x) = 1
⇒ f’\(\left(\frac{3}{2}\right)\) = 1

Question 14.
If f(x) = z cos x then f’\(\left(\frac{\pi}{2}\right)\) is equal to
1) \(\frac{-\pi}{2}\)
2) \(\frac{\pi}{2}\)
3) \(\frac{\pi}{4}\)
4) π
Answer:
1) \(\frac{-\pi}{2}\)

Explanation:
f(x) = x ⇒ cos x = f'(x) = – x sinx + cosx
⇒ f’\(\frac{-\pi}{2}\) = \(-\frac{\pi}{2}\)sin\(\frac{\pi}{2}\) + cos \(\frac{\pi}{2}=\frac{-\pi}{2}\)

Question 15.
If f(x) = 1 + x + x² + x³ + x⁴ +…. + xⁿ then f’(1) =
(1) n
(2) n + 1
(3) n(n + 1)
(4) \(\frac{\mathrm{n}+\mathrm{n}^2}{2}\)
Answer:
f(x) = 1 + x + x² + x³ + x⁴ + …….. +xⁿ
⇒ f(x) = 1 + 2x + 3x² + 4x³ + ………. + nx^n-1
⇒ f(1) = 1 + 2 + 3 + 4 + ………. + n = \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}=\frac{\mathrm{n}^2+\mathrm{n}}{2}\)

Referring to the AP Inter 1st Year Maths Study Material Chapter 11 Introduction to 3D Geometry Exercise 11a Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Introduction to 3D Geometry Solutions Exercise 11a

Question 1.
A point is on the X-axis. What are its y-coordinates and Z-coordinates.
Solution:
If a point is on the X-axis, then its y-coordinate and z-coordinates are zero.

Question 2.
A point is in the XZ-Plane. What can you say about its y-coordinate?
Solution:
If a point is in the XZ plane, then its y-coordinate is zero.

Question 3.
Name the octants in which the following points lie (1, 2, 3), (4, -2, 3), (4, -2,-5), (4, 2, -5) (-4, 2, -5), (-4, 2, 5), (-3, -1, 6), (-2, -4, -7).
Solution:
1) The x-coordinate, y-coordinate and z-coordinate of point (1, 2, 3) are all positive. Therefore, this point lies in octant I. i.e. (1, 2, 3) → I

2) The x-coordinate, y-coordinate and z-coordinate of point (4, -2, 3) are positive, negative and positive respectively. Therefore, this point lies in octant IV.
i. e., (4, -2, 3) → IV

3) The x-coordinate, y-coordinate and z-coordinate of point (4, -2, -5) are positive, negative and negative respectively. Therefore, this point lies in octant VIII.
i. e., (4, -2, -5) → VIII

4) The x-coordinate, y-coordinate and z-coordinate of point (4, 2, -5) are positive, positive and negative respectively. Therefore, this point lies in octant V.
i.e., (4, 2, -5) → V

5) The x-coordinate, y-coordinate and z-coordinate of point (-4, 2, -5) are negative, positive and negative respectively. Therefore, this point lies in octant VI.
i.e., (-4, 2, -5) → VI

6) The x-coordinate, y-coordinate and z-coordinate of point (-4, 2, 5) are negative, positive and positive respectively. Therefore, this point lies in octant II.
i.e., (-4, 2, 5) → II

7) The x-coordinate, y-coordinate and z-coordinate of point (-3, -1, 6) are negative, negative and positive respectively. Therefore, this point lies in octant III.
i.e., (-3, -1, 6) → III

8) The x-coordinate, y-coordinate and z-coordinate of point (-2, -4, -7) are all negative. Therefore, this point lies in octant VII.
i.e., (-2, -4, -7) → VII

Question 4.
Fill in the Blanks.
i) The X-axis and Y-axis taken together determine a plane known as .
ii) The coordinates of points in the XY-plane are of the form .
iii) Coordinate planes divide the space into octants.
Solution:
i) The X-axis 8s Y-axis taken together determine a plane known as XY-plane.
ii) The coordinates of points in the xy-plane are of the form (x, y, 0).
iii) Coordinate planes divide the space into eight octants.

Question 5.
i) The point (x, 0, 0) lies on _________ axis.
ii) The point (0, y, 0) lies on _________ axis.
iii) The point (0, 0, z) lies on _________ axis.
iv) The point (x, y, 0) lies on _________ plane.
v) The point (x, 0, z) lies on _________ plane.
Solution:
i) X-axis
ii) Y-axis
iii) Z-axis
iv) XY-plane
v) ZX-plane

Referring to the AP Inter 1st Year Maths Study Material Chapter 11 Introduction to 3D Geometry Exercise 11b Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Introduction to 3D Geometry Solutions Exercise 11b

I.

Question 1.
Find the distance between the following pairs of points.
(i) (2, 3, 5) and (4, 3, 1)
(ii) (-3, 7, 2) and (2, 4, -1)
(iii) (-1, 3, -4) and (1, -3, 4)
(iv) (2, -1, 3) and (-2, 1, 3).
Solution:
The distance between P(x₁, y₁, z₁) & Q(x₂, y₂, z₂) is given by
PQ = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)

i) Distance between P(2, 3, 5) & Q(4, 3, 1)
= \(\sqrt{(4-2)^2+(3-3)^2+(1-5)^2}=\sqrt{(2)^2+(0)^2+(-4)^2}=\sqrt{4+16}=\sqrt{20}\) = 2√5

ii) Distance between P(-3, 7, 2) & Q(2, 4, -1)
= \(\sqrt{(2+3)^2+(4-7)^2+(-1-2)^2}=\sqrt{(5)^2+(-3)^2+(-3)^2}=\sqrt{25+9+9}\)

iii) Distance between P(-1, 3, -4) & Q(1, -3, 4)
= \(\sqrt{(1+1)^2+(-3-3)^2+(4+4)^2}=\sqrt{(2)^2+(-6)^2+(8)^2}=\sqrt{4+36+64}=\sqrt{104}\)

iv) Distance between P(2, -1, 3) & Q(-2, 1, 3)
= \(\sqrt{(-2-2)^2+(1+1)^2+(3-3)^2}=\sqrt{(-4)^2+(2)^2+(0)^2}=\sqrt{16+4}=\sqrt{20}\) = 2√5

Question 2.
Find the distance of P(3, -2, 4) from the origin.
Solution:
Given point is P(3, -2, 4) = (x₁; y₁, z₁), Coordinates of origin ‘O’ = (0, 0, 0).
Distance OP = \(\sqrt{x_1^2+y_1^2+z_1^2}=\sqrt{9+4+16}=\sqrt{29}\)

Question 3.
Find ‘x’ if the distance between (5, -1, 7) and (x, 5, 1) is ‘9’ units.
Solution:
Given points are A(5, -1, 7), B(x, 5, 1)
⇒ Given that AB = 9
Squaring on both sides AB² = 81
⇒ (x – 5)² + (5 + 1)² + (1 – y)² = 81
⇒ (x – 5)² + 36 + 36 = 81
⇒ (x – 5)² = 81 – 72 = 9
⇒ x – 5 = ± 3
⇒ x – 5 = 3; x – 5 = -3
⇒ x = 8 (or) 2

II.

Question 1.
Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear.
Solution:
Let the points P(-2, 3, 5), Q(1, 2, 3) & R(7, 0, -1) are collinear if they lie on a same line.
PQ = \(\sqrt{(1+2)^2+(2-3)^2+(3-5)^2}=\sqrt{(3)^2+(-1)^2+(-2)^2}=\sqrt{9+1+4}=\sqrt{14}\)
QR = \(\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}=\sqrt{(6)^2+(-2)^2+(-4)^2}=\sqrt{36+4+16}=\sqrt{56}=2 \sqrt{14}\)
PR = \(\sqrt{(7+2)^2+(0-3)^2+(-1-5)^2}=\sqrt{(9)^2+(-3)^2+(-6)^2}=\sqrt{81+9+36}=\sqrt{126}=3 \sqrt{14}\)
Here, PQ + QR = \(\sqrt{14}+2 \sqrt{14}=3 \sqrt{14}\) = PR
Hence, points P(-2, 3, 5), Q(1, 2, 3) & R(7, 0, -1) are collinear.

Question 2.
Verify the following.
(i) (0, 7, -10), (1, 6, -6) and (4, 9, -6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of a right angled triangle.
(iii) (-1, 2, 1), (1, -2, 5), (4, -7, 8) and (2, -3, 4) are the vertices of a parallelogram.
Solution:
i) Let points (0, 7, -10), (1, 6, -6) & (4, 9, -6) are denoted by A, B & C respectively.
AB = \(\sqrt{(1-0)^2+(6-7)^2+(-6+10)^2}=\sqrt{(1)^2+(-1)^2+(4)^2}=\sqrt{1+1+16}=\sqrt{18}=3 \sqrt{2}\)
BC = \(\sqrt{(4-1)^2+(9-6)^2+(-6+6)^2}=\sqrt{(3)^2+(3)^2}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)
CA = \(\sqrt{(0-4)^2+(7-9)^2+(-10+6)^2}=\sqrt{(-4)^2+(-2)^2+(-4)^2}=\sqrt{16+4+16}=\sqrt{36}\) = 6
Here, AB = BC ≠ CA.
Thus, the given points are the vertices of an isosceles triangle.

ii) Let (0, 7, 10), (-1, 6, 6) & (-4, 9, 6) are denoted by A, B & C respectively.
AB = \(\sqrt{(-1-0)^2+(6-7)^2+(6-10)^2}=\sqrt{(-1)^2+(-1)^2+(-4)^2}=\sqrt{1+1+16}=\sqrt{18}\)
BC = \(\sqrt{(-4+1)^2+(9-6)^2+(6-6)^2}=\sqrt{(-3)^2+(3)^2+(0)^2}=\sqrt{9+9}=\sqrt{18}\)
CA = \(\sqrt{(0+4)^2+(7-9)^2+(10-6)^2}=\sqrt{(4)^2+(-2)^2+(4)^2}=\sqrt{16+4+16}=\sqrt{36}\)
Now AB² + BC² = 18 + 18 = 36 = CA².
Therefore, by Pythagoras theorem, ABC is a right angled triangle. Hence, the given points are the vertices of a right angled triangle.

iii) Let (-1, 2, 1), (1, -2, 5), (4, -7, 8) and (2, -3, 4) are denoted by A, B, C and D respectively.
AB = \(\sqrt{(1+1)^2+(-2-2)^2+(5-1)^2}=\sqrt{4+16+16}=\sqrt{36}\) = 6
BC = \(\sqrt{(4-1)^2+(-7+2)^2+(8-5)^2}=\sqrt{9+25+9}=\sqrt{43}\)
CD = \(\sqrt{(2-4)^2+(-3+7)^2+(4-8)^2}=\sqrt{4+16+16}=\sqrt{36}\) = 6
DA = \(\sqrt{(-1-2)^2+(2+3)^2+(1-4)^2}=\sqrt{9+25+9}=\sqrt{43}\)
Here AB = CD = 6 & BC = DA = \(\sqrt{43}\)
Hence, the opposite sides of quadrilateral ABCD are equal therefore, ABCD is a parallelogram.
Hence, the given points are the vertices of a parallelogram.

Question 3.
Find the equation of the set of points which are equidistant from the points (1. 2, 3) and (3, 2, -1).
Solution:
Let P(x, y, z) be the point that is equidistant from points A(1, 2, 3) & B(3, 2, -1). According to Geometric condition PA = PB
PA² = PB²
⇒ (x – 1)² + (y – 2)² + (z – 3)² = (x – 3)² + (y – 2)² + (z + 1)²
⇒ x² – 2x + 1 + y² – 4y + 4 + z² – 6z + 9 = x² – 6x + 9 + y² – 4y + 4 + z² + 2z + 1
⇒ -2x – 4y – 6z + 14 = -6x – 4y + 2z + 14
⇒ -2x – 6z + 6x – 2z = 0
⇒ 4x – 8z = 0
⇒ x – 2z = 0
Thus, the required equation is x – 2z = 0.

Question 4.
Find the equation of the set of points P, the sum of whose distances from A(4, 0, 0) and B(-4, 0, 0) is equal to 10.
Solution:
Let the coordinates of P be (x, y, z). The coordinates of points A & B are (4, 0, 0) & (-4, 0, 0) respectively. It is given that PA + PB = 10.
⇒ \(\sqrt{(x-4)^2+y^2+z^2}+\sqrt{(x+4)^2+y^2+z^2}\) = 10
⇒ \(\sqrt{(x-4)^2+y^2+z^2}\) = 10 – \(\sqrt{(x-4)^2+y^2+z^2}\)
Squaring on both sides, we obtain
⇒ (x – 4)² + y² + z² = 100 – \(\sqrt{(x+4)^2+y^2+z^2}\) + (x + 4)² + y² + z²
⇒ x² – 8x + 16 + y² + z² = 100 – \(\sqrt{\left(x^2+8 x+16+y^2+z^2\right.}\) + x² + 8x + 16 + y² + z²
⇒ 20 \(\sqrt{x^2+8 x+16+y^2+z^2}\) = 100 + 16x
⇒ 5 \(\sqrt{x^2+8 x+16+y^2+z^2}\) = 25 + 4x
Squaring on both sides again, we have
25(x² + 8x + 16 + y² + z²) = 625 + 16x² + 200x
⇒ 25x² + 200x + 400 + 25y² + 25z² = 625 + 16x² + 200x
⇒ 9x² + 25y² + 25z² – 225 = 0
Thus, the required equation is 9x² + 25y² + 25z² – 225 = 0.

Question 5.
If a variable point which moves such that 3PA = 2PB if A = (-2, 2, 3) and B = (13, -3, 13). Prove that P satisfies the equation x² + y² + z² + 28x – 12y + 10z – 247 = 0.
Solution:
Let P(x, y, z) be any point on the locus. Given points are A = (-2, 2, 3), B = (13, -3,’ 13) and given geometric condition is 3PA = 2PB
⇒ 9PA² = 4PB²
⇒ 9[(x + 2)² + (y – 2)² + (z – 3)²] = 4[(x – 13)² + (y + 3)² + (z – 13)²]
⇒ 9[x² + 4 + 4x + y² + 4 – 4y + z² + 9 – 6z] = 4[x² + 169 – 26x + y² + 9 + 6y + z² + 169 – 26z]
⇒ 5x² + 5y² + 5z2 + 140x – 60y + 50z – 1235 = 0
⇒ x² + y² + z² + 28x – 12y + 10z – 247 = 0.

Referring to the AP Inter 1st Year Maths Study Material Chapter 11 Introduction to 3D Geometry Exercise 11c Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Introduction to 3D Geometry Solutions Exercise 11c

I.

Question 1.
Three vertices of a parallelogram ABCD are A(3, -1, 2), B(1, 2, -4) and C(-l, 1, 2). Find the coordinates of the fourth vertex.
Solution:
Let A = (x₁, y₁, z₁) = (3, -1, 2) B = (x₂, y₂, z₂) = (1, 2, -4) C = (x₃, y₃, z₃) = (-1, 1, 2) D = (x₄, y₄, z₄) = (α, β, γ)

The fourth vertex of a parallelogram is
D = [x₁ + x₃ – x₂, y₁ + y₃ – y₂, z₁ + z₃ – z₂]
(α, β, γ) = (3 – 1 -1, -1 + 1 – 2, 2 + 2 + 4)
α = 1, β = -2, γ = 8
D = (1,-2, 8) is the fourth vertex of a parallelogram.

Question 2.
If the origin is the centroid of the triangle PQR with vertices P(2a, 2, 6), Q(-4, 3b, -10) and R(8, 14, 2c), then find the values of a, b and c.
Solution:
Let P = (x₁, y₁, z₁) = (2a, 2, 6)
Q = (x₂, y₂, z₂) = (-4, 3b, -10)
R = (x₃, y₃, z₃) = (8, 14, 2c)

Since origin is centroid then G = (0, 0, 0)
The centroid of the triangle PQR is G = \(\frac{1}{2}\)
(0, 0, 0) = \(\left[\frac{2 a-4+8}{3}, \frac{2+3 b+14}{3}, \frac{6-10+2 c}{3}\right]\)
⇒ (0, 0, 0) = \(\left(\frac{2 a+4}{3}, \frac{3 b+16}{3}, \frac{2 c-4}{3}\right)\)

∴ a = -2, b = \(\frac{-16}{3}\), c = 2

Question 3.
If A and B be the points (3, 4, 5) and (-1, 3, -7), respectively, find the equation of the set of points P such that PA² – PB² = k², where k is a constant.
Solution:
Let P(x, y, z) be the locus of a point.
Let A = (3, 4, 5) B = (-1, 3,-7)
Since PA² – PB² = k²
⇒ (x – 3)² + (y – 4)² + (z – 5)² – (x + 1)² – (y – 3)² – (z + 7)² = k²

⇒ -8x – 2y – 24z – 9 = k²
⇒ 8x + 2y + 24z + 9 + k² = 0
⇒ 8x + 2y + 24z + 9 + k² = 0 which is the required equation.

II.

Question 1.
Find the lengths of the medians of the triangle with vertices A(0, 0, 6), B(0, 4, 0) and C(6, 0, 0).
Solution:
Let A = (0, 0, 6), B = (0, 4, 0), C = (6, 0, 0)
Let D is the midpoint of \(\overline{\mathrm{BC}}\)
D = \(\left[\frac{0+6}{2}, \frac{4+0}{2}, \frac{0+0}{2}\right]\) = [3, 2, 0]

Let E is the midpoint of \(\overline{\mathrm{AC}}\)
E = \(\left[\frac{0+6}{2}, \frac{0+0}{2}, \frac{6+0}{2}\right]\) = [3, 0, 3]

Let F is the midpoint of \(\overline{\mathrm{AB}}\)
F = \(\left[\frac{0+0}{2}, \frac{0+4}{2}, \frac{6+0}{2}\right]\) = [0, 2, 3]

So \(\overline{\mathrm{AD}}, \overline{\mathrm{BE}}, \overline{\mathrm{CF}}\) will be the medians of the triangle.
Since the length of \(\overline{\mathrm{AD}}=\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2+\left(\mathrm{z}_2-\mathrm{z}_1\right)^2}\)
\(\overline{\mathrm{AD}}=\sqrt{(3-0)^2+(2-0)^2+(0-6)^2}=\sqrt{9+4+36}=\sqrt{49}\) = 7

The length of \(\overline{\mathrm{BE}}\) is \(\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2+\left(\mathrm{z}_2-\mathrm{z}_1\right)^2}\)
\(\overline{\mathrm{BE}}=\sqrt{(3-0)^2+(0-4)^2+(3-0)^2}=\sqrt{9+16+9}=\sqrt{34}\)

The length of \(\overline{\mathrm{CF}}\) is \(\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2+\left(\mathrm{z}_2-\mathrm{z}_1\right)^2}\)
\(\overline{\mathrm{CF}}=\sqrt{(6-0)^2+(0-2)^2+(0-3)^2}=\sqrt{36+4+9}=\sqrt{49}\) = 7

Question 2.
A(5, 4, 6) B(1, -1, 3) and C(4, 3, 2) are three points. Find the co-ordinates of the point in which the bisector ∠BAC meets the side BC.
Solution:
Let A = (5, 4, 6), B = (1, -1, 3), C = (4, 3, 2)
AB = \(\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2+\left(\mathrm{z}_2-\mathrm{z}_1\right)^2}\)
= \(\sqrt{(1-5)^2+(-1-4)^2+(3-6)^2}=\sqrt{16+25+9}=\sqrt{50}\) = 5√2

AC = \(\sqrt{(4-5)^2+(3-4)^2+(2-6)^2}=\sqrt{(-1)^2+(-1)^2+(-4)^2}\)
= \(\sqrt{1+1+16}=\sqrt{18}\) = 3√2
Since the angular bisector of ∠BAC divides BC in the ratio AB : AC = 5√2 : 3√2 = 5:3
Let ‘D’ divides BC in the ratio 5 : 3
D = \(\left[\frac{5(4)+3(1)}{5+3}, \frac{5(3)+3(-1)}{5+3}, \frac{5(2)+3(3)}{5+3}\right]\)
= \(\left[\frac{20+3}{8}, \frac{12}{8}, \frac{19}{8}\right]=\left[\frac{23}{8}, \frac{3}{2}, \frac{19}{8}\right]\)

Question 3.
Show that the points 0(0, 0, 0), A(2, -3, 3) and B(-2, 3, -3) are collinear. Find the ratio in which each point divides the segment joining the other two points.
Solution:
Given points are O = (0, 0, 0), A = (2, -3, 3), B = (-2, 3, -3)
Now OA = \(\sqrt{4+9+9}=\sqrt{22}\)
AB = \(\sqrt{(2+2)^2+(3+3)^2+(-3-3)^2}=\sqrt{88}=2 \sqrt{22}\)
BO = \(\sqrt{(0+2)^2+(0-3)^2+(0+3)^2}=\sqrt{22}\)
OA + BO = \(\sqrt{22}+\sqrt{22}=2 \sqrt{22}\) = AB
O, A, B are collinear points.
The ratio in which P(x, y, z) divides Q(x₁, y₁, z₁) and R(x₂, y₂, z₂) is \(\frac{x_1-x}{x-x_2}\) (or) \(\frac{y_1-y}{y-y_2}\) (or) \(\frac{z_1-z}{z-z_2}\)
Now
\(\frac{O A}{A B}=\frac{0-2}{2+2}=\frac{-2}{4}=\frac{-1}{2}\),
\(\frac{A B}{B O}=\frac{2+2}{-2-0}=\frac{4}{-2}=\frac{-2}{1}\) and
\(\frac{\mathrm{OB}}{\mathrm{OA}}=\frac{2-0}{0+2}=\frac{2}{2}=\frac{1}{1}\)

Referring to the AP Inter 1st Year Maths Study Material Chapter 11 Introduction to 3D Geometry Exercise 11d Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Introduction to 3D Geometry Solutions Exercise 11d

Question 1.
Find the coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) in the ratio (i) 2 : 3 internally, (ii) 2 : 3 externally.
Solution:
i) The coordinates of point R that divides the line segment joining the points P(x₁, y₁; z₁) and Q(x₂, y₂, z₂) internally in the ratio m : n are
\(\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}, \frac{m z_2+n z_1}{m+n}\right)\)
Let R(x, y, z) be a point that divides the line segment joining points (-2, 3, 5) and (1, -4, 6) internally in the ratio 2:3.
Hence, x = \(\frac{2(1)+3(-2)}{2+3}\), y = \(\frac{2(-4)+3(3)}{2+3}\) and z = \(\frac{2(6)+3(5)}{2+3}\)
i.e x = \(\frac{-4}{5}\),y = \(\frac{1}{5}\) and z = \(\frac{27}{5}\)
Thus, the coordinates of the required point are \(\left(\frac{-4}{5}, \frac{1}{5}, \frac{27}{5}\right)\)

ii) The coordinates of point R that divides the line segment joining the points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) externally in the ratio m : n are
\(\left(\frac{m x_2-n x_1}{m-n}, \frac{m y_2-n y_1}{m-n}, \frac{m z_2-n z_1}{m-n}\right)\)
Let R(x, y, z) be a point that divides the line segment joining points (-2, 3, 5) and (1, -4, 6)’externally in the ratio 2 : 3.
Hence, x = \(\frac{2(1)-3(-2)}{2-3}\), y = \(\frac{2(-4)-3(3)}{2-3}\) and z = \(\frac{2(6)-3(5)}{2-3}\)
i.e., x = -8, y = 17 and z = 3
Thus, the coordinates of the required point are (-8, 17, 3).

Question 2.
Given that P(3, 2, -4), Q(5, 4, -6) and R(9, 8, -10) are collinear. Find the ratio in which 9 divides PR.
Solution:
Let point Q(5, 4, -6) divides the line segment joining points P(3, 2, -4) and R(9, 8, -10) in the ratio k : 1.
Therefore, by section formula,
(5, 4, -6) = \(\frac{1}{2}\)
Hence, \(\frac{9 k+3}{k+1}\) = 5 ⇒ 9k + 3 = 5k + 5
⇒ 4k = 2
⇒ k = \(\frac{2}{4}=\frac{1}{2}\)
Thus, point Q divides PR in the ratio 1:2.

Question 3.
Find the ratio in which the YZ-plane divides the line segment formed by joining the points (-2, 4, 7) and (3, -5, 8).
Solution:
Let the YZ plane divides the line segment joining points (-2, 4, 7) and (3, -5, 8) in the ratio k : 1.
Hence, by section formula, the coordinates of point of intersection are given by
\(\left(\frac{\mathrm{k}(3)-2}{\mathrm{k}+1}, \frac{\mathrm{k}(-5)+4}{\mathrm{k}+1}, \frac{\mathrm{k}(8)+7}{\mathrm{k}+1}\right)\)
On the YZ plane, the x-coordinate of any point is zero. 3k-2
\(\frac{3 k-2}{k+1}\) = 0
⇒ 3k – 2 = 0
⇒ k = \(\frac{2}{3}\)
Thus, the YZ-plane divides the line segment formed by joining the given points in the ratio 2 : 3.

Question 4.
Using section formula, show that the points A(2, -3, 4), B(-1, 2, 1) and are collinear.
Solution:
Let P be a point that divides AB in the ratio k : 1.
Hence, by section formula, the coordinates of point of intersection are given by
\(\left(\frac{k(-1)+2}{k+1}, \frac{k(2)-3}{k+1}, \frac{k(1)+4}{k+1}\right)\)
Now, we find the value of k at which point P coincides with point C.
By taking \(\\frac{-\mathrm{k}+2}{\mathrm{k}+1}\) = 0 , we obtain k = 2.
For k = 2, the coordinates of point P are (0, \(\frac{1}{3}\), 2)
i.e., (0, \(\frac{1}{3}\), 2) is a point that divides AB internally in the ratio 2:1. Hence, points A, B and C are collinear.

Question 5.
Find the coordinates of the points which trisect the line segment joining the points P(4, 2, -6) and Q(10, -16, 6).
Solution:
Let A and B be the points that trisect the line segment joining the points P(4, 2, -6) and Q(10 -16, 6). Point A divides PQ in the ratio 1 : 2.
Therefore, by section formula, the coordinates of point A are given by
\(\left(\frac{1(10)+2(4)}{1+2}, \frac{1(-16)+2(2)}{1+2}, \frac{1(6)+2(-6)}{1+2}\right)\) = (6, -2, -2)
Point B divides PQ in the ratio 2:1.
Therefore, by section formula, the coordinates of point B are given by
\(\left(\frac{2(10)+1(4)}{1+2}, \frac{2(-16)+1(2)}{1+2}, \frac{2(6)+1(-6)}{1+2}\right)\) = (8, -10, 2)
Thus, (6, -4, -2) and (8, -10, 2) are the points that trisect the line segment joining the points P(4, 2, -6) and Q(10, -16, 6).

Practice AP Inter 1st Year Maths Study Material Chapter 11 Introduction to 3D Geometry MCQ to identify your strengths and weak areas.

AP Inter 1st Year Maths Introduction to 3D Geometry MCQ

Question 1.
The co-ordinates of the foot of the perpendicular from a point P(6, 7, 8) on the Y-axis are
1) (6, 0, 0)
2) (0, 7, 0)
3) (0, 0, 8)
4) (0, 7, 8)
Answer:
2) (0, 7, 0)

Explanation:
On Y-axis, x and z-coordinates of every point are zero so the coordinates of the foot of the perpendicular are (0, 7, 0).

Question 2.
The length of the perpendicular drawn from the point P(13, 5, 12) on the X-axis is
1) 13
2) 2\(\sqrt{13}\)
3) \(\sqrt{194}\)
4) \(\sqrt{313}\)
Answer:
1) 13

Explanation:
The foot of the perpendicular from P(13, 5, 12) on the X-axis is (13, 0, 0).
Length of the perpendicular = \(\sqrt{(4-1)^2+(5-2)^2+(8-5)^2}=3 \sqrt{3}\)

Question 3.
If a parallelepiped is formed by planes drawn through the points (1, 2, 5) and (4, 5, 8) parallel to the coordinate planes, then the length of the diagonal of the parallelepiped is
1) 2√2
2) 3
3) √3
4) 3√3
Answer:
4) 3√3

Explanation:
The length of the diagonal = \(\sqrt{(4-1)^2+(5-2)^2+(8-5)^2}\) = 3√3

Question 4.
The Locus of the point for which y = 0 is
1) XY-plane
2) YZ-plane
3) ZX-plane
4) None of these
Answer:
3) ZX-plane

Explanation:
The y-coordinate of every point of ZX-plane is zero the locus of the point (x, y, z) for which y = 0 is the XZ-plane.

Question 5.
If the distance between the points P(0, a, 3) and Q(3, 0, 7) is \41, then a =
1) -4
2) 4
3) ±4
4) 4
Answer:
3) ±4

Explanation:
PQ = \(\sqrt{41}\)
⇒ \(\sqrt{(0-3)^2+(a-0)^2+(3-7)^2}=\sqrt{41}\)
⇒ 9 + a² + 16 = 41
⇒ a² = 16
⇒ a = +4

Question 6.
The ratio in which the line joining (2, 4, 5) and (3, 5, -9) is divided by the YZ-plane
1) 2 : 3
2) 3 : 2
3) -2 : 3
4) 4 : -3
Answer:
3) -2 : 3

Explanation:
YZ-plane divides in the ratio = \(\frac{-x_1}{x_2}=\frac{-2}{3}\)

Question 7.
The points (1, 1, 1), (4, 1, 1), (4, 5, 1) and (1, 5, 1) are the vertices of ( 1 ) .
1) a rectangle
2) a square
3) parallelogram
4) None of these
Answer:
1) a rectangle

Explanation:
A = (1, 1, 1), B(4, 1, 1), C = (4, 5, 1), D = (1, 5, 1)
AB = \(\sqrt{9+0+0}\) = 3, BC = \(\sqrt{0+16+0}\) = 4, CD \(\sqrt{9+0+0}\) = 3, DA = \(\sqrt{0+16+0}\) = 4,

AC = \(\sqrt{9+16+0}\) = 5, BD = \(\sqrt{9+16+0}\) = 5
AB CD, BC = AD, AC = BD
∴ABCD is a rectangle.

Question 8.
What is the locus of a point for which y = 0, z = 0?
1) X-axis
2) Y-axis
3) Z-axis
4) YZ-plane
Answer:
X-axis

Explanation:
X-axis

Question 9.
XZ-plane divides the line joining (2, 3, 1) and (6, 7, 1) in the ratio
1) 3 : 7
2) 2 : 7
3) -3 : 7
4)-2 : 4
Answer:
3) -3 : 7

Explanation:
XZ-plane divides in the ratio = \(\frac{-y_1}{y_2}=\frac{-3}{7}\)

Question 10.
The co-ordinatea of the foot of the perpendicular drawn from the point P(3, 4, 5) on the YZ-plane are ( 2 )
1) (3, 4, 0)
2) (0, 4, 5)
3) (3, 0, 5)
4) (3, 0, 0)
Answer:
2) (0, 4, 5)

Explanation:
The point on the YZ-plane becomes its x-co-ordinate must be zero (0, 4, 5)

Question 11.
Statement-1 (Assertion) : Point (3, a² – 1, a² – 3a + 2) lies on the X-axis if a = 1
Statement-2 (Reason) : On X-axis, y and z co-ordinates of every point are equal to zero.
1) Statement-1 and statement-2 are true, statement-2 is a correct explanation for statement -1.
2) Statement-1 and statement-2 are true, statement-2 is not a correct explanation for statement-1.
3) Statement-1 is true, statement-2 is false.
4) Statement-1 is false, statement-2 is true.
Answer:
1) Statement-1 and statement-2 are true, statement-2 is a correct explanation for statement -1.

Explanation:
Statement-1 (Assertion) : If a = 1 then a² – 1 = 0, and a² – 3a + 2 = 0 the point is (3, 0, 0) lies on X-axis.
Statement-2 (Reason) : It is a correct explanation of statement-I. Hence option 1 is correct.

Referring to the AP Inter 1st Year Maths Study Material Chapter 10 Conic Sections Exercise 10a Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Conic Sections Solutions Exercise 10a

I. In each of the following exercises 1 to 5, find the equation of the circle with

Question 1.
Centre (0, 2) and radius 2.
Solution:
The equation of a circle with centre (h, k) and radius r is given as (x – h)² + (y – k)² = r². It is given that centre (h, k) = (0, 2) and radius (r) = 2 Therefore, the equation of the circle is
(x – 0)² + (y – 2)² = 2²
⇒ x² + y² + 4 – 4y = 4
⇒ x² + y² – 4y = 0

Question 2.
Centre (-2, 3) and radius 4.
Solution:
The equation of the circle with centre (h, k) and radius Y is given as (x – h)² + (y – k)² = r² .
It is given that centre (h, k) = (- 2, 3) and radius (r) = 4 Therefore, the equation of the circle is
(x + 2)² + (y – 3)² = (4)²
⇒ x² + 4x + 4 + y² – 6y + 9 = 16
⇒ x² + y² + 4x – 6y – 3 = 0.

Question 3.
Centre \(\left(\frac{1}{2}, \frac{1}{4}\right)\) and radius \(\frac{1}{12}\).
Solution:
The equation of the circle with centre (h, k) and radius r is given as (x – h)² + (y – k)² = r²
It is given that centre (h, k) = \(\left(\frac{1}{2}, \frac{1}{4}\right)\) and radius (r) = \(\frac{1}{12}\)

Therefore, the equation of the circle is (x – \(\frac{1}{2}\))² + (y – \(\frac{1}{4}\))² = \(\left(\frac{1}{12}\right)^2\)
⇒ x² – x + \(\frac{1}{4}\) + y² – \(\frac{y}{2}+\frac{1}{16}=\frac{1}{144}\)
⇒ 144x² – 144x + 36 + 144y² – 72y + 9 – 1 = 0 .
⇒ 144x² – 144x + 144y² – 72y + 44 = 0
⇒ 36x² – 36x + 36y² – 8y + 11 = 0
⇒ 36x² + 36y² – 36x – 18y + 11 = 0.

Question 4.
Centre (1, 1) and radius √2.
Solution:
The equation of the circle with centre (h, k) and radius r is given as (x – h)² + (y – k)² = r²
It is given that centre (h, k) = (1, 1) and radius (r) = 7²
Therefore, the equation of the circle is (x – 1)² + (y – 1)² = (72)²
⇒ x² – 2x + 1 + y² – 2y + 1 = 2
⇒ x² + y² – 2x – 2y = 0.

Question 5.
Centre (-a, -b) and radius \(\sqrt{a^2-b^2}\).
Solution:
The equation of the circle with centre (h, k) and radius r is given as (x – h)² + (y – k)² = r²
It is given that centre (h, k) = (-a, -b) and radius (r) = \(\sqrt{a^2-b^2}\)
Therefore, the equation of the circle is (x + a)² + (y + b)² = \(\left(\sqrt{a^2-b^2}\right)^2\)
⇒ x² + 2ax + a² + y² + 2by + b² = a² – b²
x² + y² + 2ax + 2by + 2b² = 0

In each of the following Exercises 6 to 10, find the centre and radius of the circles.

Question 6.
(x + 5)² + (y – 3)² = 36.
Solution:
The equation of the given circle is (x + 5)² + (y – 3)² = 36
⇒ [x – (-5)]² + (y – 3)² = 6²
which is in the form of (x – h)² + (y – k)² = r², where h = – 5, k = 3 and r = 6
Thus, the centre of the given circle is (-5, 3), while its radius is 6.

Question 7.
x² + y² – 4x – 8y – 45 = 0
Solution:
The equation of the given circle is x² + y² – 4x – 8y – 45 = 0
⇒ (x² – 4x) + (y² – 8y) = 45
⇒ [x² – 2(x)(2) + 2²] + [y² – 2(y) (4) + 4²] – 4 – 16 = 45
⇒ (x – 2)² + (y – 4)² = 65
⇒ (x – 2)² + (y – 4)² = \((\sqrt{65})^2\)
Which is in the form of (x – h)² + (y – k)² = r², where h = 2, k = 4 and r = \(\sqrt{65}\)
Thus, the centre of the given circle is (2, 4), while its radius is \(\sqrt{65}\)

(OR)

Let S = x² + y² – 4x – 8y – 45 = 0
Comparing to x² + y² + 2gx + 2fy + c = 0
2g = -4
⇒ g = -2; 2f = -8
⇒ f = -4 ; c = – 45

i) Centre C = (-g, -f) = (2, 4)
ii) Radius r = \(\sqrt{g^2+f^2-c}=\sqrt{4+16+45}=\sqrt{65}\)

Question 8.
x² + y² – 8x + 10y – 12 = 0
Solution:
The equation of the given circle is x² + y² – 8x + 10y – 12 = 0
⇒ (x² – 8x) + (y² + 10y) = 12
⇒ [x² – 2(x) (4) + 4²] + [y² + 2(y) (5) + (5²)] – 16 – 25 = 12
⇒ (x – 4)² + (y + 5)² = 53
⇒ (x – 4)² + [y – (-5)]² = \((\sqrt{53})^2\)
Which is in the form of (x – h)² + (y – k)² = r² where h = 4, k = -5 and r = \(\sqrt{53}\)
Thus, the centre of the given circle is (4, -5) and its radius is \(\sqrt{53}\)

Question 9.
2x² + 2y² – x = 0
Solution:
The equation of the given circle is 2x² + 2y² – x = 0
⇒ (2x² – x) + 2y² = 0
⇒ 2[(x² – \(\frac{1}{4}\)) + y²] = 0
⇒ [x² – 2x(\(\frac{1}{4}\)) + \(\left(\frac{1}{4}\right)^2\)] = 0
⇒ (x – \(\frac{1}{4}\))² + (y – 0)² = \(\left(\frac{1}{4}\right)^2\)
which is in the form of (x – h)² + (y – k)² = r², where h = \(\frac{1}{4}\), k = 0 and r = \(\frac{1}{4}\)
Thus, the centre of the circle is (\(\frac{1}{4}\), 0), while its radius is \(\frac{1}{4}\).

Question 10.
\(\sqrt{1+\mathrm{m}^2}\)(x² + y²) – 2cx – 2mcy = 0.
Solution:
\(\sqrt{1+\mathrm{m}^2}\)(x² + y²) – 2cx – 2mcy = 0
⇒ x² + y² – \(\frac{2 \mathrm{c}}{\sqrt{1+\mathrm{m}^2}}\)x – \(\frac{2 \mathrm{mc}}{\sqrt{1+\mathrm{m}^2}}\)y = 0

2g = \(\frac{-2 \mathrm{c}}{\sqrt{1+\mathrm{m}^2}}\), 2f = \(\frac{-2 \mathrm{mc}}{\sqrt{1+\mathrm{m}^2}}\), c = 0

⇒ C = (-g, -f) = \(\left(\frac{c}{\sqrt{1+m^2}}, \frac{m c}{\sqrt{1+m^2}}\right)\)
Radius = \(\sqrt{g^2+f^2-c}=\sqrt{\frac{c^2}{1+m^2}+\frac{m^2 c^2}{1+m^2}}\) = c units

Question 11.
Find the equation of the circle passing through the origin and having the centre at (- 4, – 3).
Solution:
Circle passes through the origin and having the centre (-4, -3). i.e., (h, k) = (-4, -3); (x, y) = (0, 0)
Equation of the circle is (x – h)² + (y – k)² = r²
⇒ (0 + 4)² + (0 + 3)² = r²
⇒ 16 + 9 = r²
⇒ r² = 25
The required equation of the circle, (x + 4)² + (y + 3)² = 25
⇒ x² + y² + 8x + 6y = 0.

Question 12.
Find the equation of the circle passing through (2, -1) having the centre at (2, 3).
Solution:
C = (2, 3), P = (2, -1),
adius CP = \(\sqrt{(2-2)^2+(3+1)^2}\) = 4
Equation of the circle is (x – h)² + (y – k)² = r²
⇒ (x – 2)² + (y – 3)² = 4²
x² + y² – 4x – 6y – 3 = 0

Question 13.
Find the equation of the circle passing through (-2, 3), having the centre at (0, 0).
Solution:
C = (0, 0), P = (-2, 3), Radius = \(\sqrt{(0+2)^2+(0-3)^2}=\sqrt{13}\)
∴ Equation of the circle is (x – 0)² + (y – 0)² = \((\sqrt{13})^2\)
⇒ x² + y² = 13.

Question 14.
Find the equation of the circle passing through (3, 4) and having the centre at (-3, 4).
Solution:
(h, k) = (-3, 4); (x, y) = (3, 4)
Equation of the circle is (x – h)² + (y – k)² = r²
(x + 3)² + (y – 4)² = r²
⇒ (3 + 3)² + (4 – 4)² = r²
⇒ r² = 36
The required equation of the circle is (x + 3)² + (y – 4)² = 36
= x² + 6x + 9 + y² – 8y + 16 – 36 = o
x² + y² + 6x – 8y – 11 = 0

Question 15.
Find the value of ‘a’ If 2x² + ay² – 3x + 2y – 1 = 0 represents a circle and also find its radius.
Solution:
The general equation of second degree equation ax² + 2hxy + by² + 2gx + 2fy + c = 0, represents a circle if a = b, h = 0, g² + f² – c ≥ 0.
2x² + ay² – 3x + 2y – 1 = 0 represents a circle, then a = 2.
= x² + y² – x + y = 0
2g = \(\frac{3}{2}\), 2f = 1, c = –\(\frac{1}{2}\)
= C = (-g, -f) = \(\left(\frac{3}{4},-\frac{1}{2}\right)\)
Radius = \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{\frac{9}{16}+\frac{1}{4}+\frac{1}{2}}=\frac{\sqrt{21}}{4}\) units

Question 16.
Find the values of a&b if ax² + bxy + 3y² – 51 + 2y – 3 = 0 repyesentgacfrcle.
Also find the radius and centre of the circle.
Solution:
The general equation of second degree ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a circie, then a = b, h = 0.
∴ ax² + bxy + 3y² – 5x + 2y – 3 = 0 represents a circle.
b = 0, a = 3
3x² + 3y² – 5x + 2y – 3 = 0
⇒ x² + y² – \(\frac{5}{3}\)x + \(\frac{2}{3}\)y – 1 = 0
⇒ g = \(\frac{-5}{6}\), f = \(\frac{2}{6}=\frac{1}{3}\), c = -1
⇒ C = (-g, -f) = \(\left(\frac{5}{6},-\frac{1}{3}\right)\)
Radius = \(\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{\frac{9}{16}+\frac{1}{4}+\frac{1}{2}}=\frac{\sqrt{21}}{4}\) units.

Question 17.
If x² + y² + 2gx + 2fy – 12 = 0 represents a circle with centre (2, 3) find g, f and its radius.
Solution:
Given circle is x² + y² + 2gx + 2fy – 12 = 0,
g = -2, f = -3, c = -12, Radius = \(\sqrt{g^2+f^2-c}=\sqrt{4+9+12}\) = 5 units.

Question 18.
If x² + y² + 2gx + 2fy = 0 represents a circle with centre (-4, -3) then find g, f and the radius of the circle.
Solution:
Given circle is x² + y² + 2gx + 2fy = 0, C = (-g, -f) = (-4, -3)
g = 4, f = 3
Radius = \(\sqrt{g^2+f^2-c}=\sqrt{16+9}\) = 5 units.

Question 19.
If x² + y² – 4x + 6y + c = 0 represents a circle with radius 6 then find the value of c.
Solution:
Given circle is x² + y² – 4x + 6y + c = 0, g = – 2, f = 3.
Radius = \(\sqrt{g^2+f^2-c}\) = 6
\(\sqrt{4+9-c}\) = 6
⇒ 13 – c = 36
⇒ c = -23

Question 20.
Find the equation of a circle which Is concentric with x² + y² – 6x – 4y – 12 = 0 and passing through (-2, 14).
Solution:
Given circle S = x² + y² – 6x – 4y – 12 = 0, Let P = (-2, 14)
The concentric circle S = x² + y² – 6x – 4y + k = 0 ………. (1)
Which passing through (-2, 14)
⇒ (-2)² + (14)² – 6(2) – 4(14) + k = 0
⇒ 4 + 196 + 12 – 56 + k = 0
⇒ 156 + k = 0
⇒ k = -156
∴ The concentric circle S’ = x² + y² – 6x – 4y – 156 = –

Question 21.
Solution:
The centre of the circle is given as (h, k) = (2, 2)
Since the circle .passes through the point (4, 5) and radius (r) of the circle is the distance between the points (2, 2) and (4, 5)
r = \(\sqrt{(2-4)^2+(2-5)^2}=\sqrt{(-2)^2+(-3)^2}=\sqrt{4+9}=\sqrt{13}\)
The equation of the circle is (x – h)² + (y – k)² = r²
⇒ (x – 2)² + (y – 2)² = \((\sqrt{13})^2\)
⇒ x² – 4x + 4 + y² – 4y + 4 = 13
⇒ x² + y² – 4x – 4y – 5 = 0.

Question 22.
Does the point (-2.5, 3.5) lie inside, outside or on the circle x² + y² = 25 ?
Solution:
The equation of the given circle is x² + y² = 25
⇒ (x – 0)² + (y – 0)² = 5², which is in the form of (x – h)² + (y – k)² = r²
Here, h = 0, k = 0, r = 5
∴ centre = (0, 0) and radius = 5
Distance between the point (-2.5, 3.5) and centre (0, 0)
= \(\sqrt{(-2.5-0)^2+(3.5-0)^2}=\sqrt{6.25+12.25}=\sqrt{18.5}\) = 4.3 < 5
Hence the point (-2.5, 3.5) lies inside the circle.

II.

Question 1.
If the abscissae of points A, B are the roots of the equation x² + 2ax – b² = 0 and ordinates of A, B are roots of y² + 2py – q² = 0 then find the equation of a circle for which AB is a diameter.
Solution:
If A (x₁, y₁) and B (x₂, y₂) are the extremities of diameter AB, then the equation of circle is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
⇒ x² – x(x₁ + x₂) + x₁x₂ + y² – y (y₁ + y₂) + y₁y₂ = 0
x₁ x₂ are the roots of x² + 2ax – b² = 0; y₁, y₂ are the roots of y² + 2py – q² = 0
⇒ x₁ + x₂ = – 2a;
⇒ x₁x₂ = – b²;

⇒ y₁ + y₂ = -2p
⇒ y₁y₂ = -q²
Equation of the circle is x² – x(-2a) – b² + y² – y(- 2p) – q² = 0
x² + y² + 2ax + 2py – b² – q² = 0

Question 2.
Show that A(3, -1) lies on the circle x² + y² – 2x + 4y = 0. Also find the other end of the diameter through A.
Solution:
Equation of the circle is x² + y² – 2x + 4y = 0 ………. (1)
A (3, -1) and Let B(x₁, y₁),
Substitute A in equation (1), we get
(3)² + (-1)² – 2(3) + 4(-1) = 0
⇒ ∴ A lies on the circle, C(-g, -f) = (1, -2)
C is the centre of the circle and also C is the midpoint of AB.
\(\frac{x_1+3}{2}\) = 1 ⇒ x₁ = -1; \(\frac{y_1-1}{2}\) = -2
⇒ y₁ = -3
∴ B(x₁, y₁) = (-1, 3)

Question 3.
Show that A(-3, 0) lies on x² + y² + 8x + 12y + 15 = 0 and find the other end of the diameter through A.
Solution:
Equation of the circle is x² + y² + 8x + 12y + 15 = 0 …… (1)
Substitute A (-3, 0) in equation (1), we get (-3)² + (0)² + (-3) + 12(0) + 15 = 0
A(-3, 0) is one end of the diameter, Centre is C(-4, -6).
Let B(x₁, y₁) be the other end. And also ‘C’ is the mid point of AB. then,
\(\frac{x_1+(-3)}{2}\) = -4
⇒ x₁ = -5; \(\frac{y_1-0}{2}\) = -6
⇒ y₁ = -12
∴ Other end of the diameter is (-5, -12).

Question 4.
Find the equation of the circle whose centre lies on X-axis and passing through (-2, 3) and (4, 5).
Solution:
Let the equation of circle is x² + y² + 2gx + 2fy + c = 0 ………….. (1)
(-2, 3) and (4, 5) passing through (1)
4 + 9 – 4g + 6f + c = 0 ……………… (2)
16 + 25 + 8g+ 10f+ c = 0 …………. (3)
Equation (3) – (2), we get, 28 + 12g + 4f = 0 ⇒ f + 3g = -7
Centre lies on X – axis , f = 0 ⇒ g = \(\frac{-7}{3}\), f = 0, c = \(\frac{-67}{3}\)
Substitute g, f and c values in equation (1).
The required equation of the circle be
x² + y² + 2(\(\left(-\frac{7}{3}\right)\))x + 2(0)y – \(\frac{67}{3}\) =0
⇒ 3 (x² + y²) – 14x – 67 = 0.

III.

Question 1.
Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Solution:
The equation of the required circle be (x – h)² + (y – k)² = r² since the circle passes through the points (4, 1) and (6, 5)
⇒ (4 – h)² + (1 – k)² = r² …………. (1)
and (6 – h)² + (5 – k)² = r² ………… (2)
since the centre (h, k) of the circle lies on the line 4x + y = 16,
4h + k = 16 ………….. (3)
from equations (1) and (2)
(4 – h)² + (1 – k)² = (6 – h)² + (5 – k)²
⇒ 16 – 8h + h² + 1 – 2k + k² = 36 – 12h + h² + 25 – 10k + k²
⇒ 16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k
⇒ 4h + 8k = 44
⇒ h + 2k = 11 ……………. (4)
On solving equations (3) and (4) we have h = 3 and k = 4
Substituting the values of h and k in equation (1) is
⇒ (4 – 3)² + (1 – 4)² = r²
⇒ (1)² + (- 3)² = r²
⇒ 1 + 9 = r²
⇒ r² = 10
⇒ r = Jio
Thus the equation of the required circle is
(x – 3)² + (y – 4)² = \(\sqrt{10}\)
⇒ x² – 6x + 9 + y² – 8y + 16 = 10
⇒ x² + y² – 6x – 8y + 15 = 0

Question 2.
Find the equation of the circle passing through the points (2, 3) and (-1. 1) and whose centre is on the line x – 3y – 11 = 0.
Solution:
The equation of the required circle be (x – h)² + (y – k)² = r²
Since the circle passes through the points (2, 3) and (-1, 1),
(2 – h)² + (3 – k)² = r² ………. (1)
and (-1 – h)² + (1 – k)² = r² ………….. (2)
Since the centre (h, k) of the circle lies on the line x – 3y – 11 = 0,
∴ h – 3k = 11 ………… (3)
from equations (1) and (2) (2 – h)² + (3 – k)² = (-1 – h)² + (1 – k)²
⇒ 4 – 4h + h² + 9 – 6k + k² = 1 + 2h + h² + 1 – 2k + k²
⇒ 4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k
⇒ 6h + 4k = 11 ………… (4)
By solving equations (3) and (4) we have, h = \(\frac{7}{2}\) and k = \(\frac{-5}{2}\)
Substituting h, k values in (1) ⇒ (2 – \(\frac{7}{2}\))² + (3 + \(\frac{5}{2}\))² = r²
⇒ \(\left(\frac{4-7}{2}\right)^2+\left(\frac{6+5}{2}\right)^2\) = r²
⇒ \(\left(\frac{-3}{2}\right)^2+\left(\frac{11}{2}\right)^2\) = r²
⇒ \(\frac{9}{4}+\frac{121}{4}\) = r²
⇒ \(\frac{130}{4}\) = r²

Thus the equation of the required circle is (x – \(\frac{7}{2}\))² + (y + \(\frac{5}{2}\))² = \(\frac{130}{2}\)
⇒ \(\frac{1}{2}\)
⇒ 4x² + 4y² – 28x + 20y – 56 = 0
⇒ x² + y² – 7x + 5y – 14 = 0

Question 3.
Find the equation of a circle which passes through the points (2, -3) and (-4, 5) and having the centre on 4x + 3y + 1 = 0.
Solution:
Given points are A(2, – 3), B(- 4, 5). Given line is 4x + 3y + 1 = 0
Equation of the circle is x² + y² + 2gx + 2fy + c = 0 ………….. (1)
A(2, -3) ⇒ 4 + 9 + 4g – 6f + c = 0 ⇒ 4g – 6f + c + 13 = 0 …………. (2)
C(-g, – f) lies on the given line, – 4g – 3f + 1 = 0 ………… (4)
(2) and (3)

3g – 4f = 7 ………. (5)
Now solving (4) & (5)

\(\frac{g}{21+4}=\frac{f}{3-28}=\frac{1}{16+9}\)
⇒ \(\frac{g}{25}=\frac{f}{-25}=\frac{1}{25}\)
g = 1; f = -1 ⇒ (g, f) = (1, -1)
(2) ⇒ 4(1) + (- 6) (-1) + c = -13
⇒ 4 + 6 + c = -13
⇒ c = – 23
Substituting g, f, c values in (1) ⇒ x² + y² + 2(1)x + 2(-1)y – 23 = 0
x² + y² + 2x – 2y – 23 = 0

Question 4.
Find the equation of a circle which passes through (4, 1), (6, 5) and having the centre on 4x + 3y – 24 = 0.
Solution:
Equation of the circle be x² + y² + 2gx + 2fy + c = 0 passes through (4, 1) and (6, 5) then
4² + 1² + 2g(4) + 2f(1) + c = 0 ………… (1)
6² + 5² + 2g(6) + 2f(5) + c = 0 ………… (2)
Centre lies on 4x + 3y – 24 = 0
⇒ 4 (-g) + 3(-f) – 24 = 0 ………….. (3)
(2) – (1), we get 44 + 4g + 8f = 0 …………. (4)
Solving (3) and (4) we get f = -4, g = -3, c = 15
∴ Required equation of the circle is x² + y² – 6x – 8y + 15 = 0.

Question 5.
Find the equation of the circle with radius 5 whose centre lies on the x-axis and passes through the point (2, 3).
Solution:
The equation of the required circle be (x – h)² + (y – k)² = r2
Since the radius of the circle is 5, and its centre lies on the X-axis, k = 0 and r = 5. Now, the equation of the circle becomes (x – h)² + y² = 25
It is given that the circle passes through the point (2, 3).
Therefore, (2 – h)² + 3² = 25
⇒ (2 – h)² = 16
⇒ 2 – h = ±4
If 2 – h = 4 then h = – 2 and
If 2 – h = – 4 then h = 6
when h = – 2, the equation of the circle becomes (x + 2)² + y² = 25
⇒ x² + 4x + 4 + y² = 25
⇒ x² + y² + 4x – 21 = 0.
when h = 6, the equation of the circle becomes (x – 6)² + y² = 25
⇒ x² – 12x + 36 + y² = 25
⇒ x² + y² – 12x + 11 = 0.

Question 6.
Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.
Solution:
Let the equation of the required circle is (x – h)² + (y – k)² = r²
Since the centre of the circle passes through (0, 0)
Therefore (0 – h)² + (0 – k)² = r²
⇒ h² + k² = r²
The equation of the circle becomes (x – h)² + (y – k)² = h² + k²
It is given that the circle makes intercepts a and b on the coordinate axes. This
means that the circle passes through the points (a, 0) and (0, b).
(a – h)² + (0 – k)² = h² + k² ………. (1),
(0 – h)² + (b – k)² = h² + k² ………. (2)
From equation (1) we have a² – 2ah + h² + k² = h² + k²
⇒ a² – 2ah = 0
⇒ a(a – 2h) = 0 a = 0 (or) (a – 2h) = 0
However, a ≠ 0, hence, (a – 2h) = 0
h² + b² – 2bk + k² = h² + k²
⇒ b² – 2bk = 0
⇒ b(b – 2k) = 0
⇒ b = 0 (or) (b – 2k) = 0
However, b ≠ 0; hence (b – 2k) = 0
A (0, 0) lies on S = 0 ⇒ c = 0

iii) (2, 1), (5, 5), (-6, 7).
Solution:
(2, 1), (5, 5), (-6, 7)
Squation of the circle is x² + y² + 2gx + 2fy + c = 0
(2, 1), (5, 5), (-6, 7) satisfy the above equation, 4 + 1 + 4g + 2f + c = 0 ……….. (1)
25 + 25 + 10g + 10f + c = 0 …………….(2)
36 + 49 – 12g + 14f + c = 0 ………. (3)
Subtract (2) – (1), we get, 45 + 6g + 8f = 0 ……….. (4)
Subtract (3) – (2), we get, 35 – 22g + 4f = 0 ………… (5)
Solve (4) and (5), we get, g = ½, f = -6, c = 5
The required equation of circle x² + y² + x – 12y + 5 = 0

Referring to the AP Inter 1st Year Maths Study Material Chapter 10 Conic Sections Exercise 10b Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Conic Sections Solutions Exercise 10b

I. In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

Question 1.
y² = 12z
Solution:
The given equation is y² = 12x
Here, the coefficient of x is positive, hence the parabola opens towards the right.
On comparing this equation with y² = 4ax,
we have 4a = 12 ⇒ a = 3
Coordinates of the focus = (a, 0) = (3, 0)
Since the given equation involves y², the axis of the parabola is the X-axis.
Equation of directrix, x = – a, i.e., x = -3 ⇒ x + 3 = 0
Length of the latus rectum = 4a = 4 × 3 = 12.

Question 2.
x² = 6y
Solution:
The given equation is x² = 6y
Here, the coefficient ofy is positive. Hence, the parabola opens upwards on comparing the equation with x² = 4ay
4a = 6 ⇒ a = \(\frac{3}{2}\)
⇒ Coordinates of the focus = (0, a) = (0, \(\frac{3}{2}\))
Since the given equation involves x², the axis of the parabola is the Y-axis (x = 0).
Equation of directrix y = \(\frac{-3}{2}\). Length of the latus rectum = 4a = 6

Question 3.
y² = -8x
Solution:
The given equation is y² = -8x
Here, the coefficient of “x’ is negative. Hence, the parabola opens towards left. On comparing this equation with y² = – 4ax
– 4a = -8
⇒ a = 2
Coordinates of the focus = (-a, 0) = (- 2, 0).
Since the given equation involves y², the axis of the parabola is the X-axis (y = 0). Equation of the directrix, x = a i.e., x = 2, Length of the latus rectum = 4a = 8.

Question 4.
x² = -16y
Solution:
The given equation is x² = -16y
Here, the coefficient of y is negative, hence the parabola opens downwards. On comparing this equation with x² = – 4ay, we have -4a = -16
⇒ a = 4
∴ The coordinates of the focus = (0, -a) = (0, -4)
Since the equation involves x², the axis of the parabola is Y-axis (x = 0).
Equation of directrix, y = a i.e., y = 4
Length of the latus rectum = 4a = 16

Question 5.
y² = 10x
Solution:
The given equation is y² = 10x
Here, the coefficient of x is positive, hence the parabola opens towards the right.
On comparing this equation with y² = 4ax,
4a = 10 ⇒ a = \(\frac{5}{2}\),
∴ Coordinates of the focus = (a, 0) = (\(\frac{5}{2}\), 0)
Since the given equation involves y², the axis of the parabola is the X-axis (y = 0).
Equation of the directrix is x = -a i.e., x = \(\frac{-5}{2}\). Length of the latus rectum = 4a = 10

Question 6.
x² = – 9y
Solution:
The given equation is x² = – 9y
Here, the coefficient of y is negative. Hence, the parabola opens downwards.
On comparing this equation with x² = – 4ay,
-4a = – 9
a = \(\frac{9}{4}\)
∴ Coordinates of the focus = (0, -a) = (0, \(\frac{-9}{4}\))
Since, the equation involves x², the axis of the parabola is the Y-axis (x = 0).
Equation of the directrix y = a i.e., y = \(\frac{9}{4}\). Length of the latus rectum = 4a = 9.

In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions.

Question 7.
Focus (6, 0); directrix x = – 6
Solution:
Focus (6, 0); directrix, x = – 6
Since the focus lies on the X-axis, the X-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form y² = 4ax or y² = -4ax
It is also seen that the directrix, x = – 6 is to the left of the Y-axis, while the focus (6, 0) is right of the Y-axis.
Hence, the parabola is in the form of y² = 4ax with a = 6
Thus, the equation of the parabola is y² = 24x.

Question 8.
Focus (0, – 3); directrix y = 3
Solution:
Focus = (0, – 3); directrix y = 3
Since the focus lies on Y-axis, the Y-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form x² = 4ay or x² = -4ay.
It is also seen that the directrix, y = 3 is above the X-axis, while the focus (0, -3) is below the X-axis.
Hence, the parabola is of the form x² = -4ay with a = 3
Thus, the equation of the parabola is x² = -12y.

Question 9.
Vertex (0, 0); focus (3, 0)
Solution:
Vertex (0, 0); focus (3, 0)
Since the vertex of the parabola is (0, 0), and the focus lies on the positive X-axis.
Here, X-axis is the axis of the parabola, while the equation of the parabola is of the form y² = 4ax. Since the focus is (3, 0), a = 3
Thus, the equation of the parabola is y² = 4 × 3 × x
⇒ y² = 12x.

Question 10.
Vertex (0, 0); focus (-2, 0)
Solution:
Vertex (0, 0); focus (-2, 0)
Since the vertex of the parabola is (0, 0), and the focus lies on the negative X-axis.
Here, X-axis is the axis of the parabola, while the equation of the parabola is of the
form y² = 4ax. Since the focus is (-2, 0), a = 2
Thus, the equation of the parabola is y² = – 4 × —2 × x
⇒ y² = 8x.

Question 11.
Vertex (0. 0) passing through (2, 3) and axis is along X-axls.
Solution:
Since the vertex is (0, 0) and the axis of the parabola is the X-axis, the equation of the parabola is either of the form y² = 4ax or y² = – 4ax.
The parabola passes through the point (2, 3), which lies in the first quadrant.
So, the equation of the parabola is of the form y² = 4ax, while (2, 3) must satisfy the equation y² = 4ax,
∴ 3² = 4a(2)
a = \(\frac{9}{8}\)
Thus, the equation of the parabola is y² = 4\(\left(\frac{9}{8}\right)\)x
⇒ 2y² = 9x

Question 12.
Vertex (0, 0), passing through (5, 2) and symmetric with respect to Y-axis.
Solution:
Since the vertex is (0, 0) and the parabola is symmetric about the Y-axis, the equation of the parabola is either of the form x² = 4ay or x² = – 4ay.
The parabola passes through the point (5, 2), which lies in the first quadrant.
So, the equation of the parabola is of the form x² = 4ay, while (5, 2) must satisfy the equation x² = 4ay
∴ (5)² = 4 × a × 2
⇒ 25 = 8a
⇒ a = \(\frac{25}{8}\)

II.

Question 1.
Find the equation of the parabola whose focus is S(3, 5) and vertex is A(1, 3).
Solution:
Given S = (3, 5), A =(1, 3).
Equation of the axis is y – 3 = \(\frac{3-5}{1-3}\)(x – 1)
⇒ y – 3 = x – 1
⇒ x – y + 2 = 0
The directrix is perpendicular to axis. Equation of the directrix isx + y + k = 0
Let the co-ordinates of Z be (x, y). A is the midpoint of SZ. Then,
\(\frac{3+x}{2}\) = 1 ⇒ 3 + x = 2
⇒ x = -1; \(\frac{5+y}{2}\) = 3
⇒ 5 + y = 6
⇒ y = 1.
Co-ordinates of Z are (-1, 1).
The directrix passes through Z (-1, 1) ⇒ -1 + 1 + k = 0 ⇒ k = 0
Equation of the directrix is x + y = 0
Equation of the parabola is (x – α)² + (y + β)² = \(\frac{(l x+m y+n)^2}{l^2+m^2}\)
(x – 3)² + (y – 5)² = \(\frac{(x+y)^2}{1+1}\)
⇒ 2(x² – 6x + 9 + y² – 10y + 25) = (x + y)²
⇒ 2x² + 2y² – 12x – 20y + 68 = x² + 2xy + y²
⇒ x² – 2xy + y² – 12x – 20y + 68 = 0

Question 2.
Find the equation of the parabola whose focus is (-2, 3) and directrix is the line 2x + 3y – 4 = 0. Also find the length of the latus rectum and the equation of the axis of the parabola.
Solution:
Assume P (x₁, y₁) is any point on the parabola, S(-2, 3) is the focus.
⇒ SP² = (x₁ + 2)² + (y₁ – 3)².
Equation of the directrix is 2x + 3y – 4 = 0.
PM is the perpendicular from P on the directrix PM = \(\frac{\left|2 x_1+3 y_1-4\right|}{\sqrt{4+9}}\)
SP = PM
⇒ SP² = PM²
⇒ (x₁ + 2)² + (y₁ – 3)² = \(\frac{1}{2}\)
⇒ 13(x₁² + 4x₁ + 4 + y₁² – 6y₁ + 9) = (2x₁ + 3y₁ – 4)²
⇒ 13x₁² + 13y₁² + 52x₁ – 78y₁ + 169) = 4x₁² + 9y₁² + 16 + 12x₁y₁ – 16x₁ – 24y₁
⇒ 9x₁² – 12x₁y₁ + 4y₁² + 68x₁ – 54y₁ + 153 = 0
Locus of P(x₁, y₁) is 9x² – 12xy + 4y² + 68x – 54y + 153 = 0
Length of the latus rectum = 4a
Perpendicular distance from S on directrix = \(\frac{|2(-2)+3(3)-4|}{\sqrt{4+9}}=\frac{1}{\sqrt{13}}\)
2a = \(\frac{1}{\sqrt{13}}\) ⇒ 4a = \(\frac{2}{\sqrt{13}}\). The axis is perpendicular to the directrix.
Equation of the directrix can be taken as 3x – 2y + k = 0.
The line passes through S (-2, 3) ⇒ -6 – 6 + k = 0
⇒ k = 12
Equation of the axis is 3x – 2y + 12 = 0.

Question 3.
Find the equation of the parabola whose axis is parallel to X-axis and which passes through the points (-2, 1), (1, 2) and (-1, 3).
Solution:
Axis is parallel to X – axis, x = ay² + by + c.
Passes through (-2, 1), (1, 2), (-1, 3)
-2 = a + b + c ………… (1)
1 = 4a + 2b + c ……….. (2)
– 1 = 9a + 3b + c ……… (3)
(2) – (3), we get 2 = – 5a – b …….. (4)
(2) – (1), we get 3 = 3a + b …… (5)
(4) + (5), we get 5 = – 2a ⇒ a = \(\frac{-5}{2}\) ⇒ b = \(\frac{21}{2}\),c = -10
x = \(\frac{-5}{2}\)y² + \(\frac{21}{2}\)y – 10
⇒ 5y² + 2x – 21y + 20 = 0.

Question 4.
Find the equation of the parabola whose axis is parallel to Y-axis and which passes through the points (4, 5), (-2, 11) and (-4, 21).
Solution:
Axis is parallel to Y-axis, y = ax² + bx + c.
Passes through (4, 5), (-2, 11), (-4, 21)
5 = 16a + 4b + c ……….. (1)
11 = 4a – 2b + c ……….. (2)
21 = 16a – 4b + c ……..(3)
(2) – (1), we get 6 = -12a – 6b ….. (4)
(2) – (3), we get 10 = 12a – 2b ……. (5)
Solve (4) and (5), we get a = \(\frac{1}{2}\)
⇒ b = -2, c = 5
y = \(\frac{1}{2}\)x² – 2x + 5
⇒ x² – 2y – 4x + 10 = 0.

Referring to the AP Inter 1st Year Maths Study Material Chapter 10 Conic Sections Exercise 10c Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Conic Sections Solutions Exercise 10c

I.

Question 1.
Find the eccentricity of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1.
Solution:
Given ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 (a < b).
Comparing to \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
a² = 9, b² = 16
Eccentricity (e) = \(\sqrt{\frac{b^2-a^2}{b^2}}=\sqrt{\frac{16-9}{16}}=\frac{\sqrt{7}}{4}\)

Question2.
Find the length of the latus rectum of ellipse x² + 16y² = 16
Solution:
Given ellipse is x² + 16y² = 16 ⇒ \(\frac{x^2}{16}+\frac{y^2}{1}\) = 1 (a > b)
Comparing to \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1, a² = 16, b² = 1
The length of latus rectum = \(\frac{2 b^2}{a}=\frac{2(1)}{4}=\frac{1}{2}\)

II. In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

Question 1.
\(\frac{x^2}{36}+\frac{y^2}{16}\) = 1
Solution:
The given equation is \(\frac{x^2}{36}+\frac{y^2}{16}\) = 1
Here, the denominator of \(\frac{x^2}{36}\) is greater than the denominator of \(\frac{y^2}{16}\).
Therefore, the major axis is along the X-axis, while the minor axis is along the Y-axis.
On comparing the given equation with \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1. Now, a = 6 and b = 4
∴ c = \(\sqrt{a^2-b^2}=\sqrt{36-16}=\sqrt{20}\) = 2√5
The coordinates of the foci are (2√5,0) and (- 2√5,0)
The coordinates of the vertices are (6, 0) and (- 6, 0)
Length of the major axis = 2a = 12, Length of the minor axis = 2b = 8
Eccentricity, e = \(\frac{c}{a}=\frac{2 \sqrt{5}}{6}=\frac{\sqrt{5}}{3}\). Length of the latus rectum = \(\frac{2 b^2}{a} \Rightarrow \frac{2 \times 16}{6}=\frac{16}{3}\).

Question 2.
\(\frac{x^2}{4}+\frac{y^2}{25}\) = 1
Solution:
The given equation is \(\frac{x^2}{4}+\frac{y^2}{25}\) = 1
⇒ \(\frac{x^2}{2^2}+\frac{y^2}{5^2}\) = 1
Here, the denominator of \(\frac{y^2}{25}\) is greater than the denominator of \(\frac{x^2}{4}\)
Therefore, the major axis along the Y-axis, while the minor axis is along the X-axis.
On comparing the given equation with \(\frac{x^2}{b^2}+\frac{y^2}{a^2}\)j = 1,
b = 2 and a = 5 ⇒ c = \(\sqrt{a^2-b^2}=\sqrt{25-4}=\sqrt{21}\)
Therefore, the coordinates of foci are (0, \(\sqrt{21}\)) and (0, –\(\sqrt{21}\))
The coordinates of the vertices are (0, 5) and (0, -5)
Length of the major axis = 2a = 10, Length of the minor axis = 2b = 4
Eccentricity, e = \(\sqrt{a^2-b^2}=\sqrt{25-4}=\sqrt{21}\), Length of the latus rectum

Question 3.
\(\frac{x^2}{16}+\frac{y^2}{9}\) = 1
Solution:
The given equation is \(\frac{x^2}{16}+\frac{y^2}{9}\) = 1
⇒ \(\frac{x^2}{4^2}+\frac{y^2}{3^2}\) = 1
Here, the denominator of \(\frac{\mathrm{x}^2}{16}\) is greater than the denominator of \(\frac{y^2}{9}\)
Therefore, the major axis along the X-axis, while the minor axis is along the Y-axis.
On comparing the given equation with \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1,
a = 4 and b = 3 ⇒ c = \(\sqrt{a^2-b^2}=\sqrt{16-9}=\sqrt{7}\)
Therefore, the coordinates of foci are (±√7,0)
The coordinates of the vertices are (± 4, 0)
Length of the major axis = 2a = 8
Length of the minor axis = 2b = 6
e = \(\frac{c}{a}=\frac{\sqrt{7}}{4}\), Length of the latus rectum = \(\frac{2 b^2}{a}=\frac{2 \times 9}{4}=\frac{9}{2}\)

Question 4.
\(\frac{x^2}{25}+\frac{y^2}{100}\) = 1
Solution:
The given equation is \(\frac{x^2}{25}+\frac{y^2}{100}\) = 1
⇒ \(\frac{x^2}{5^2}+\frac{y^2}{10^2}\) = 1
Hence, the denominator of \(\frac{y^2}{100}\) is greater than the denominator of \(\frac{x^2}{25}\)
Therefore, the major axis is along the Y-axis, while the minor axis is along the X-axis.
On comparing the given equation with \(\frac{x^2}{b^2}+\frac{y^2}{a^2}\) = 1. Now b = 5 and a = 10
∴ c = \(\sqrt{a^2-b^2}=\sqrt{100-25}=\sqrt{75}\) = 5√3
The coordinates of foci are (0, ±5√3)
Therefore, the coordinates of the vertices are (0, ±10)
Length of the major axis = 2a = 20,
Length of the minor axis = 2b = 10
e = \(\frac{c}{a}=\frac{5 \sqrt{3}}{10}=\frac{\sqrt{3}}{2}\), Length of the latus rectum = \(\frac{2 b^2}{a}=\frac{2 \times 25}{10}\) = 5

Question 5.
\(\frac{x^2}{49}+\frac{y^2}{36}\) = 1
Solution:
The given equation is \(\frac{x^2}{49}+\frac{y^2}{36}\) = 1
⇒ \(\frac{x^2}{7^2}+\frac{y^2}{6^2}\) = 1
Hence, the denominator of \(\frac{x^2}{49}\) is greater than the denominator of \(\frac{y^2}{36}\)
Therefore, the major axis is along the X-axis, while the minor axis is along the Y-axis.
On comparing the given equation with \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1,
a = 7 and b = 6
⇒ c = \(\sqrt{a^2-b^2}=\sqrt{49-36}=\sqrt{13}\)
Therefore, the coordinates of foci are (±\(\sqrt{13}\), 0)
The coordinates of the vertices are (±7, 0)
Length of the major axis = 2a = 14,
Length of the minor axis = 2b = 12
e = \(\frac{c}{a}=\frac{\sqrt{13}}{7}\), Length of the latus rectum = \(\frac{2 b^2}{a}=\frac{2 \times 36}{7}=\frac{72}{7}\)

Question 6.
\(\frac{x^2}{100}+\frac{y^2}{400}\) = 1
Solution:
The given equation is \(\frac{x^2}{100}+\frac{y^2}{400}\) = 1
⇒ \(\frac{x^2}{7^2}+\frac{y^2}{6^2}\) = 1
Here, the denominator of \(\frac{x^2}{49}\) is greater than the denominator of \(\frac{y^2}{36}\)
Therefore, the major axis is along the Y-axis while the minor axis is along the X-axis.
On comparing the given equation with \(\frac{1}{2}\) = 1
⇒ b = 10 and a = 20
c = \(\sqrt{a^2-b^2}=\sqrt{400-100}=\sqrt{300}\) = 10√3
Therefore, the coordinates of foci are (0, ±10√3)
The coordinates of vertices are (0, ±20)
Length of the major axis = 2a = 40, Length of the minor axis = 2b = 20
e = \(\frac{c}{a}=\frac{10 \sqrt{3}}{20}=\frac{\sqrt{3}}{2}\), Length of the latus rectum \(\frac{2 b^2}{a}=\frac{2 \times 100}{20}\) = 10

Question 7.
36x² + 4y² = 144
Solution:
The given equation is 36x² + 4y² = 144 can be written as
\(\frac{x^2}{4}+\frac{y^2}{36}\) = 1 ⇒ \(\frac{x^2}{2^2}+\frac{y^2}{6^2}\) = 1
Here, the denominator of \(\frac{y^2}{6^2}\) is greater than the denominator of \(\frac{x^2}{2^2}\)
Therefore, the major axis is along Y-axis while the minor axis is along the X-axis on.
On comparing equation (1) with \(\frac{x^2}{b^2}+\frac{y^2}{a^2}\) = 1
⇒ b = 2 and a = 61
c = \(\sqrt{a^2-b^2}=\sqrt{36-4}=\sqrt{32}\) = 4√2
Therefore, the coordinates of foci are (0, +4^2 ) The coordinates of the vertices are (0, ± 6) Length of the major axis = 2a = 12
Length of the minor axis = 2b = 4
e = \(\frac{c}{a}=\frac{4 \sqrt{2}}{6}=\frac{2 \sqrt{2}}{3}\),
Length of the latus rectum = \(\frac{2 b^2}{a}=\frac{2 \times 4}{6}=\frac{4}{3}\)

Question 8.
16x² + y² = 16
Solution:
The given equation is 16x² + y² = 16
It can be written by \(\frac{x^2}{1}+\frac{y^2}{16}\) = 1
⇒ \(\frac{x^2}{1^2}+\frac{y^2}{4^2}\) = 1 …. (1)
Here, the denommator of \(\frac{y^2}{4^2}\) is greater than the denominator of \(\frac{x^2}{1^2}\)
Therefore the major axis is along the Y-axis, while the minor axis is along the X-axis.
On comparing equation (1) with \(\frac{x^2}{b^2}+\frac{y^2}{a^2}\) = 1
⇒ a = 4 and b = 1
c = \(\sqrt{a^2-b^2}=\sqrt{16-1}=\sqrt{15}\) ⇒ c = \(\sqrt{15}\)
Therefore the coordinates of foci are (0, ± \(\sqrt{15}\)), the coordinate of the vertices are (0, ± 4)
Length of major axis = 2a = 8, Length of minor axis = 2b = 2
Eccentricity e = \(\frac{c}{a}=\frac{\sqrt{15}}{4}\), Length of latus rectum = \(\frac{2 b^2}{a}=\frac{2 \times 1}{4}=\frac{2}{4}=\frac{1}{2}\)

Question 9.
4x² + 9y² = 36
Solution:
The equation is 4x² + 9y² = 36
⇒ \(\frac{4 x^2}{36}+\frac{9 y^2}{36}\) = 1

⇒ \(\frac{x^2}{9}+\frac{y^2}{4}\) = 1

⇒ \(\frac{x^2}{3^2}+\frac{y^2}{2^2}\) = 1 ……………….. (1)

Here, the denominator of \(\frac{x^2}{3^2}\) is greater than the denominator of \(\frac{y^2}{2^2}\)
The major axis is along the X-axis while the minor axis along the Y-axis.
On comparing the given equation with \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1, we have a = 3 and b = 2
⇒ c = \(\sqrt{a^2-b^2}=\sqrt{9-4}=\sqrt{5}\)
The coordinates of the foci are (±√5, O)
The coordinates of the vertices are (± 3, 0)
Length of major axis = 2a = 6, Length of minor axis = 2b = 4
Eccentricity e = \(\frac{c}{a}=\frac{\sqrt{5}}{3}\), Length of the latus rectum = \(\frac{2 b^2}{a}=\frac{2 \times 4}{3}=\frac{8}{3}\)

Question 10.
Find the equation of the ellipse with focus at (1, -1), e = 2/3 and directrix as x + y + 2 = 0.
Solution:
Given focus S = (1, -1), Eccentricity e = \(\frac{2}{3}\)
Equation of directrix is x + y + 2 = 0.

Let P(x, y) be the point on the ellipse, then \(\frac{\mathrm{SP}}{\mathrm{PM}}\) = e, SP = ePM.
⇒ SP² = e²PM²
⇒ (x – 1)² + (y + 1)²2 = \(\frac{4}{9} \frac{|x+y+2|^2}{2}\)
⇒ 9[(x – 1)² + (y + 1)²] = 2(x + y + 2)²
⇒ 9[x² + 1 – 2x + y² + 1 + 2y] = 2(x² + y² + 4 + 2xy + 4y +4x)
⇒ 9x² + 9y² – 18x + 18y + 18 = 2x² + 2y² + 4xy + 8x + 8y + 8
⇒ 7x² – 4xy + 7y² – 26x + 10y + 10 = 0
∴ The equation of the ellipse is 7x² + 7y² – 4xy – 26x + 10y + 10 = 0

Question 11.
Find the equation of the ellipse in the standard form whose distance between foci is 2 and the length of latus rectum is 15/2.
Solution:
Let \(\frac{1}{2}\) = 1 ………. (1) be the equation of ellipse.
Given the distance between foci = 2
2ae = 2 ⇒ ac = 1 ⇒ Length of the latus rectum = \(\frac{15}{2}\)
\(\frac{2 b^2}{a}=\frac{15}{2}\)
⇒ b = \(\frac{15 a}{4} \Rightarrow \frac{15 a}{4}\) = a² – 1
[∵ b² = a²(1 – e²) = b² = a² – a²e²]
⇒ 15a = 4a² – 4
⇒ 4a² – 15a – 4 = 0
⇒ 4a² – 16a + a – 4 = 0
⇒ 4a (a – 4) + 1 (a – 4) = 0
⇒ (a – 4) (4a + 1) = 0
⇒ a – 4 = 0
⇒ a = 4; 4a + 10 ⇒ 4a = -1
⇒ a = –\(\frac{1}{4}\)
[∵ a = 4(+Ve)]
b² = \(\frac{15(4)}{4}\) = 15
⇒ The equation of the ellipse is \(\frac{x^2}{16}+\frac{y^2}{15}\) = 1

Question 12.
Find the equation of the ellipse In the standard form such that distance between foci is 8 and distance between directrices is 32.
Solution:
Let the equation of the ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 (a> b)
Given that the distance between foci = 8
∴ 2ae = 8 ⇒ ae = 4………. (1)
The distance between directrices = 32
\(\frac{2a}{e}\) = 32 ⇒ \(\frac{a}{e}\) = 16 ………. (2)
From (1) × (2)
⇒ ae × \(\frac{a}{e}\) = 4 × 16
⇒ a² = 64
⇒ a = 8

We Know that b² = a²(1 – e²)
b² = a² – a²e²
= (8)² – (4)²
= 64 – 16
= 48

∴ b² = 48
⇒ The equation of ellipse is \(\frac{x^2}{64}+\frac{y^2}{48}\) = 1

In each of the following BZerciaea 13 to 23, find the equation for the ellipse that satisfies the given conditions.

Question 13.
Vertices (± 5, 0), foci (± 4, 0)
Solution:
Vertices (± 5, 0), foci (± 4, 0).
Here the vertices are on the X-axis.
The equation at the ellipse will be of the form
\(\frac{1}{2}\) = 1, where ‘a is the semi-major axis
Accordingly a = 5 and c = 4. It is known that a² = b² + c²
5² = b² + 4²
25 = b² + 16 ⇒ b² = 9 ⇒ b = 3
Thus the equation of the ellipse is \(\frac{x^2}{5^2}+\frac{y^2}{3^2}\) = 1 (or) \(\frac{x^2}{25}+\frac{y^2}{9}\) = 1

Question 14.
Vertices (0, ± 13), foci (0, ± 5)
Solution:
Vertices (0, + 13), foci (0, ± 5)
Here the vertices are on the Y-axis.
So, the equation of the ellipse will be of the form \(\frac{1}{2}\) = 1, where ‘a’ is the semi¬major axis.
Accordingly a = 13 and c = 5, it is known that a² = b² + c²
⇒ 13² = b² + 5²
⇒ 169 = b² + 25
⇒ b² = 144
⇒ b = 12
Thus the equation of the ellipse is \(\frac{x^2}{12^2}+\frac{y^2}{13^2}\) = 1 (or) \(\frac{x^2}{144}+\frac{y^2}{169}\) = 1

Question 15.
Vertices (±6, 0), foci (±4, 0)
Solution:
Vertices (± 6, 0), foci (+ 4, 0), here the vertices are on the X-axis.
∴ The equation of the ellipse will be of the form “r + ^ = 1 where ‘a’ is the semi¬major axis.
Accordingly a = 6, c = 4, it is known that
a² = b² + c²
⇒ 6² = b² + 4²
⇒ 36 = b² + 16
⇒ b² = 20
⇒ b = \(\sqrt{20}\)
The equation of the ellipse is \(\frac{x^2}{3^2}+\frac{y^2}{2^2}\) = 1 ⇒ \(\frac{x^2}{9}+\frac{y^2}{4}\) = 1

Question 16.
End, of major axIs (±3, 0), ends of minor axIs (0, ± 2)
Solution:
Ends of major axis (±3, 0), ends of minor axis (0, ± 2)
The major axis is along with X-axis
∴ The equation of the ellipse will be of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 where ‘a’ is the semi- major axis.
Accordingly a = 3 and b = 2
Thus the equation of the ellipse is \(\frac{x^2}{3^2}+\frac{y^2}{2^2}\) = 1
⇒ \(\frac{x^2}{9}+\frac{y^2}{4}\) = 1

Question 17.
Ends of major axis (0, ± √5), ends of minor axis (± 1, 0)
Solution:
Ends of major axis (0, ±√5), ends of minor axis (± 1, 0)
Here, the major axis is along the Y-axis
∴ The equation of the ellipse will be of the form \(\frac{1}{2}\) = 1 where ‘a’ is semi-major
axis.
Accordingly a = √5 and b = 1
The equationof the ellipse is \(\frac{x^2}{1^2}+\frac{y^2}{(\sqrt{5})^2}\) = 1
⇒ \(\frac{x^2}{1}+\frac{y^2}{5}\) = 1.

Question 18.
Length of major axis 26. foci (± 5, 0).
Solution:
Length of major axis = 26, foci = (± 5, 0)
Since the foci are on X-axis, the major axis is along the X-axis
∴ The equation of the ellipse will be of the form + 1 where ‘a’ is the semi-major axis.
Accordingly 2a = 26 ⇒ a = 13 and c = 5
It is known that a² = b² + c²
⇒ (13)² = b² + 5²
⇒ 169 = b² + 25
⇒ b² = 144
⇒ b = 12
The equation of the ellipse is \(\frac{x^2}{13^2}+\frac{y^2}{12^2}\) = 1 ⇒ \(\frac{x^2}{169}+\frac{y^2}{144}\) = 1

Question 19.
Length of minor axis 16, foci (0, ± 6).
Solution:
Length of minor axis = 16, foci (0, ± 6)
Since the foci are on the Y-axis, the major axis is along the Y-axis.
∴ The equation of the ellipse will be of the form \(\frac{x^2}{b^2}+\frac{y^2}{a^2}\) = 1 where ‘a’ is the semi-major axis.
Accordingly 2b = 16 ⇒ b = 8 and c = 6.
∴ a² = b² + c²
⇒ a² = 8² + 6²
⇒ a² = 64 + 36
⇒ a² = 100
⇒ a = 10
The equation of the ellipe is \(\frac{x^2}{8^2}+\frac{y^2}{10^2}\) = 1 ⇒ \(\frac{x^2}{64}+\frac{y^2}{100}\) = 1.

Question 20.
Foci(±3, 0), a = 4
Solution:
Foci (±3, 0), a = 4.
Since the foci are on the X-axis the major axis is along the X-axis.
∴ The equation of ellipse will be of the form \(\frac{1}{2}\) = 1 where ‘a’ is the semi-major axis. Accordingly a = 4 and c = 3
⇒ a² = b² + c²
⇒ 4² = b² + 3²
⇒ b² = 16 – 9
⇒ b = 27
∴ Equation of the ellipse is \(\frac{x^2}{16}+\frac{y^2}{7}\) = 1

Question 21.
b = 3, c = 4, centre at the origin foci on the X-axis.
Solution:
It is given that b = 3 and c = 4 centre at the origin, foci on the X-axis.
Since the foci are on the X-axis, the major axis is along the X-axis.
∴ The equation of the ellipse will be of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 where ‘a’is the semi – major axis.
Accordingly b = 3, c = 4, it is known that
a² = b² + c²
⇒ a² = 9 + 16
⇒ a² = 25
⇒ a = 5
Equation of the ellipse is \(\frac{x^2}{5^2}+\frac{y^2}{3^2}\) = 1
\(\frac{x^2}{25}+\frac{y^2}{9}\) = 1

Question 22.
Centre at (0. 0), major axis on the Y-axis and passes through the points (3, 2) and (1, 6).
Solution:
As the centre at (0, 0) and the major axis on the Y-axis.
The equation of the ellipse will be of the form \(\frac{x^2}{b^2}+\frac{y^2}{a^2}\) = 1 …………(1)
a is the semi-major axis.
The ellipse passes through the points (3, 2) and (1, 6)
∴ \(\frac{9}{b^2}+\frac{4}{a^2}\) = 1 ………..(2)
\(\frac{1}{b^2}+\frac{36}{a^2}\) = 1 ………(3)
Solving (2) and (3) b² = 10, a² = 40
The equation of the ellipse is \(\frac{x^2}{10}+\frac{y^2}{40}\) = 1 ⇒ 4x² + y² = 40

Question 23.
Major axis on the X-axis and passes through the points (4, 3) and (6, 2).
Solution:
Since the major axis is on the X-axis the equation of the ellipse will be of the form
\(\frac{1}{2}\) = 1 ………..(1)
where a is the semi-major axis.
The ellipse passes through the points (4, 3) and (6, 2)
∴ \(\frac{16}{a^2}+\frac{9}{b^2}\) = 1 …….(2)
\(\frac{36}{a^2}+\frac{4}{b^2}\) = 1 …….(3)
Solving (2) and (3) ⇒ a² = 5² and b² = 13
Thus, the equation of the ellipse is \(\frac{x^2}{52}+\frac{y^2}{13}\) = 1 ⇒ x² + 4y² = 52