Students must practice these Maths 1B Important Questions TS Inter 1st Year Maths 1B Tangent and Normal Important Questions to help strengthen their preparations for exams.

## TS Inter 1st Year Maths 1B Tangent and Normal Important Questions

Question 1.

Find the slope of the tangent to the curve y = 5x^{2} at the point (-1, 5). [May ’04]

Solution:

Given, the equation of the curve is y = 5x^{2}

Differentiating on both sides with respect to ‘x’

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 5(2x) = 10x

∴ Slope of the tangent at P(-1, 5) is m = \(\left(\frac{d y}{d x}\right)_P\)

= 10(-1)

= -10

Question 2.

Find the equations of the tangent and the normal to the curve y = 5x^{4} at the point (1, 5). [May ’15 (AP), ’10, ’07]

Solution:

Given, the equation of the curve is y = 5x^{4}

Differentiating on both sides with respect to ‘x’

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 5 . 4x^{3} = 20x^{3}

Slope of tangent at P(1, 5) is m = \(\left(\frac{d y}{d x}\right)_P\)

= 20(1)^{3}

= 20

The equation of the tangent at P is y – y_{1} = m(x – x_{1})

y – 5 = 20(x – 1)

y – 5 = 20x – 20

20x – y – 15 = 0

The equation of normal at P is y – y_{1} = \(\frac{-1}{m}\) (x – x_{1})

y – 5 = \(\frac{-1}{20}\) (x – 1)

20y – 100 = -x + 1

x + 20y – 101 = 0

Question 3.

Find the equations of the tangent and the normal to the curve y^{4} = ax^{3} at (a, a). [May ’13]

Solution:

Given the equation of the curve is y^{4} = ax^{3}

Differentiating on both sides with respect to ‘x’

Equation of tangent at P(a, a) is y – y_{1} = m(x – x_{1})

y – a = \(\frac{3}{4}\)(x – a)

4y – 4a = 3x – 3a

3x – 4y + a = 0

Equation of normal at P (a, a) is y – y_{1} = \(\frac{-1}{m}\) (x – x_{1})

y – a = \(\frac{-1}{\frac{3}{4}}\) (x – a)

3y – 3a = 4x + 4a

4x + 3y – 7a = 0

Question 4.

II Show that the tangent at any point ‘θ’ on the curve x = c sec θ, y = c tan θ is y sin θ = x – c cos θ. [Mar. ’19 (TS)]

Solution:

Given that x = c sec θ

Differentiating on both sides with respect to ‘θ’.

\(\frac{d x}{d \theta}\) = c sec θ tan θ

and y = c tan θ

Differentiating on both sides with respect to ‘θ’

\(\frac{d y}{d \theta}\) = c sec^{2}θ

Now,

y sin θ cos θ – c sin^{2}θ = x cos θ – c

y sin θ cos θ – c sin^{2}θ + c = x cos θ

y sin θ cos θ + c(1 – sin^{2}θ) = x cos θ

y sin θ cos θ + c cos^{2}θ = x cos θ

y sin θ + c cos θ = x

y sin θ = x – c cos θ

Question 5.

Find the equations of tangent and the normal to the curve y = x^{2} – 4x + 2 at the point (4, 2).

Solution:

Given equation of the curve is y = x^{2} – 4x + 2 ……….(1)

Let given point be P(x, y) = (4, 2)

Differentiating (1) on both sides with respect to ‘x’

\(\frac{d y}{d x}\) = 2x – 4

Slope of the tangent at P is m = \(\left(\frac{\mathrm{d} y}{\mathrm{dx}}\right)_{\mathrm{P}}\)

m = 2(4) – 4 = 4

The equation of the tangent at P is y – y_{1} = m(x – x_{1})

y – 2 = 4(x – 4)

y – 2 = 4x – 16

4x – y – 14 = 0

Equation of normal at P is y – y_{1} = \(\frac{-1}{m}\) (x – x_{1})

y – 2 = \(\frac{-1}{m}\) (x – 4)

4y – 8 = -x + 4

x + 4y – 12 = 0

Question 6.

Show that the length of the subnormal at any point on the curve y^{2} = 4ax is a constant. [Mar. ’13 (Old), ’11, ’05; May ’09]

Solution:

Given, the equation of the curve is y^{2} = 4ax

Differentiating on both sides with respect to ‘x’

∴ The length of the subnormal at any point (x, y) on the curve is a constant.

Question 7.

Show that the length of the subtangent at any point on the curve y = a^{x} (a > 0) is a constant.

Solution:

Given, the equation of the curve is y = a^{x}

Differentiating on both sides with respect to ‘x’

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = a^{x} log a

The slope of the tangent at any point P(x, y) is

∴ The length of the subtangent at any point P(x, y) on the curve is a constant.

Question 8.

Show that the square of the length of sub tangent at any point on the curve by^{2} = (x + a)^{3} (b ≠ 0) varies with the length of the subnormal at that point.

Solution:

Given, equation of the curve is by^{2} = (x + a)^{3}

Differentiating on both sides with respect to ‘x’

∴ (length of subtangent)^{2} ∝ (length of the subnormal).

Question 9.

Find the lengths of sub tangent and subnormal at a point on the curve y = b sin(\(\frac{x}{a}\)). [Mar. ’16 (TS); May ’05]

Solution:

Given equation of the curve is y = b sin(\(\frac{x}{a}\))

Differentiating on both sides with respect to ‘x’

Question 10.

Show that at any point (x, y) on the curve y = \(b e^{x / a}\), the length of the subtangent is a constant and the length of the subnormal is \(\frac{\mathbf{y}^2}{a}\). [Mar. ’18 (TS); May ’12; Mar. ’10]

Solution:

Given equation of the curve is y = \(b e^{x / a}\)

Differentiating on both sides with respect to ‘x’

Question 11.

Find the lengths of normal and subnormal at a point on the curve y = \(\frac{a}{2}\left[e^{\frac{x}{a}}+e^{\frac{-x}{a}}\right]\). [Mar. ’13]

Solution:

Given equation of the curve is y = \(\frac{a}{2}\left[e^{\frac{x}{a}}+e^{\frac{-x}{a}}\right]\)

y = a cosh(\(\frac{x}{a}\))

Differentiating on both sides with respect to ‘x’

Question 12.

Show that the tangent at P(x_{1}, y_{1}) on the curve √x + √y = √a is \(\mathrm{y}_1 \mathrm{y}^{-1 / 2}+\mathrm{xx}_1^{-1 / 2}\) = \(a^{1 / 2}\). [May ’15 (AP), ’04]

Solution:

Given the equation of the curve is √x + √y = √a ……….(1)

Since the point P(x_{1}, y_{1}) lies on the curve (1) then √x_{1} + √y_{1} = √a ……….(2)

Differentiating (1) on both sides with respect to ‘x’

Which is a required equation of a tangent.

Question 13.

If the tangent at any point on the curve \(x^{2 / 3}+y^{2 / 3}=a^{2 / 3}\) intersects the coordinate axes in A and B, then show that the length AB is a constant. [Mar. ’19 (TS); Mar. ’15 (TS), ’14, ’13, ’08, ’07, ’05, ’03; May ’10]

Solution:

Given equation of the curve is \(x^{2 / 3}+y^{2 / 3}=a^{2 / 3}\)

Differentiating on both sides with respect to ‘x’

AB^{2} = a^{2}

⇒ AB = a (constant)

∴ The length AB is a constant.

Question 14.

If the tangent at any point P on the curve x^{m} y^{n} = a^{m+n} (mn ≠ 0) meets the coordinate axes in A, B, then show that AP : BP is a constant. [Mar. ’10; May ’08, ’06]

Solution:

Given the equation of the curve is x^{m} y^{n} = a^{m+n}

Differentiating on both sides with respect to ‘x’

Question 15.

At any point, ‘t’ on the curve x = a (t + sin t), y = a(1 – cos t), find the lengths of a tangent, normal, subtangent, and subnormal. [Mar. ’18 (AP)]

Solution:

Given that x = a(t + sin t)

Differentiating on both sides with respect to ‘t’

Question 16.

(i) Find the angle between the curves xy = 2 and x^{2} + 4y = 0. [Mar. ’17 (AP); May ’13, ’11]

(ii) Define the angle between the curves. [May ’11]

Solution:

(i) Given, the equations of the curves are

xy = 2 …….(1)

x^{2} + 4y = 0 ………(2)

From (1), y = \(\frac{2}{x}\)

Substituting in equation (2)

(ii) Angle between two curves: If two curves intersect at a point P and the tangents at P for both curves exists then the angle between the tangents at P is called the angle between the curves at P.

Question 17.

Show that the curves y^{2} = 4(x + 1) and y^{2} = 36(9 – x) intersect orthogonally. [Mar. ’13 (old), ’11, ’09, ’08, ’06, ’02; May ’15 (AP), ’05]

Solution:

Given curves are

y^{2} = 4(x + 1) ………(1)

y^{2} = 36(9 – x) ……..(2)

From (1) and (2)

4(x + 1) = 36(9 – x)

x + 1 = 81 – 9x

10x = 80

x = 8

If x = 8,

y^{2} = 4(8 + 1) = 4(9) = 36

y = ±6

∴ Required points are P(8, 6), Q(8, -6)

Differentiating (1) on both sides with respect to ‘x’

Now, m_{1}m_{2} = \(\frac{-1}{3}\)(3) = -1

∴ m_{1}m_{2} = -1

Since, m_{1}m_{2} = -1, the two curves cut each other orthogonally.

Question 18.

Find the angle between the curves y^{2} = 4x and x^{2} + y^{2} = 5. [Mar. ’16 (AP), ’12; May ’07]

Solution:

Given, the equations of the curves are

y^{2} = 4x ……..(1)

x^{2} + y^{2} = 5 ……..(2)

Substitute equation (1) with equation (2)

x^{2} + 4x = 5

x^{2} + 4x – 5 = 0

x^{2} + 5x – x – 5 = 0

x(x + 5) – 1(x + 5) = 0

(x + 5) (x – 1) = 0

x + 5 = 0 (or) x – 1 = 0

x = -5 (or) x = +1

If x = -5, from (1)

y^{2} = 4(-5)

y^{2} = -20 ∉ R

If x = 1, from (1)

y^{2} = 4(1)

y^{2} = 4

y = √4 = ±2

∴ Required point are P(1, 2), Q(1, -2)

Question 19.

Find the angle between the curves x^{2} + 3y = 3 and x^{2} – y^{2} + 25 = 0. [May ’03]

Solution:

Given, the equations of the curves are

x^{2} + 3y = 3 …….(1)

x^{2} – y^{2} + 25 = 0 …….(2)

From (1),

x^{2} + 3y = 3

x^{2} = 3 – 3y

Substituting in equation (2)

3 – 3y – y^{2} + 25 = 0

-y^{2} – 3y + 28 = 0

y^{2} + 3y – 28 = 0

y^{2} + 7y – 4y – 28 = 0

y(y + 7) – 4(y + 7) = 0

(y + 7)(y – 4) = 0

y + 7 = 0 (or) y – 4 = 0

y = -7 (or) y = 4

If y = -7,

x^{2} = 3 – 3(-7)

= 3 + 21

x^{2} = 24

x = ±2√6

If y = 4,

x^{2} = 3 – 3(4)

= 3 – 12

= -9 ∉ R

∴ Required points are P(2√6, -7), Q(-2√6, -7)

Differentiating (1) on both sides with respect to ‘x’

Question 20.

Find the angle between the curves 2y^{2} – 9x = 0 and 3x^{2} + 4y = 0 (in the 4th quadrant). [May ’09]

Solution:

Given, the equations of the curves are

2y^{2} – 9x = 0 ……..(1)

3x^{2} + 4y = 0 ………(2)

From (1), 2y^{2} = 9x

x = \(\frac{2 y^2}{9}\)

Substituting in equation (2)

Question 21.

Find the angle between the curves y^{2} = 8x and 4x^{2} + y^{2} = 32. [Mar. ’18 (AP); May ’12]

Solution:

Given, the equations of the curves are

y^{2} = 8x ……..(1)

4x^{2} + y^{2} = 32 ………(2)

Substituting y^{2} = 8x in equation (2)

4x^{2} + 8x = 32

4x^{2} + 8x – 32 = 0

x^{2} + 2x – 8 = 0

x^{2} + 4x – 2x – 8 = 0

x(x + 4) – 2(x + 4) = 0

(x + 4) (x – 2) = 0

x + 4 = 0 (or) x – 2 = 0

x = -4 (or) x = 2

If x = -4,

y^{2} = 8(-4) = -32 ∉ R

If x = 2,

y^{2} = 8(2) = 16

y = ±4

∴ Required point are P(2, 4), Q(2, -4).

Differentiating (1) on both sides with respect to ‘x’

2y . \(\frac{d y}{d x}\) = 8

y . \(\frac{d y}{d x}\) = 4

\(\frac{d y}{d x}=\frac{4}{y}\)

Differentiating (2) on both sides with respect to ‘x’

4(2x) + 2y . \(\frac{d y}{d x}\) = 0

8x + 2y . \(\frac{d y}{d x}\) = 0

Question 22.

Show that the curves 6x^{2} – 5x + 2y = 0 and 4x^{2} + 8y^{2} = 3 touch each other at \(\left(\frac{1}{2}, \frac{1}{2}\right)\). [Mar. ’15 (AP), ’10; May ’13 (old)]

Solution:

Given, the equations of the curves are

6x^{2} – 5x + 2y = 0 ……….(1)

4x^{2} + 8y^{2} = 3 ………(2)

Let the given point P = \(\left(\frac{1}{2}, \frac{1}{2}\right)\)

Differentiating (1) on both sides with respect to ‘x’

∴ m_{1} = m_{2}

∴ The curves 6x^{2} – 5x + 2y = 0, 4x^{2} + 8y^{2} = 3 touch each other.

### Some More Maths 1B Tangent and Normal Important Questions

Question 23.

Find the equations of the tangents to the curve y = 3x^{2} – x^{3}, where it meets the X-axis.

Solution:

Given, equation of the curve is y = 3x^{2} – x^{3} ………(1)

Given that the curve meets at X-axis then y = 0

∴ 3x^{2} – x^{3} = 0

x^{2}(3 – x) = 0

x^{2} = 0 or 3 – x = 0

x = 0 or x = 3

∴ P(0, 0), Q(3, 0)

Differentiating (1) on both sides with respect to ‘x’.

\(\frac{d y}{d x}\) = 3(2x) – 3x^{2} = 6x – 3x^{2}

at P(0, 0):

Slope of the tangent at P(0, 0) is

m = \(\left(\frac{d y}{d x}\right)_P\)

= 6(0) – 3(0)^{2}

= 0

Equation of tangent at P(0, 0) is y – y_{1} = m(x – x_{1})

y – 0 = 0(x – 0)

y = 0

at Q(3, 0):

Slope of tangent at Q(3, 0) is m = \(\left(\frac{d y}{d x}\right)_Q\)

= 6(3) – 3(3)^{2}

= 18 – 27

= -9

∴ Equation of the tangent at Q(3, 0) is y – y_{1} = m(x – x_{1})

y – 0 = -9(x – 3)

y = -9x + 27

9x + y – 27 = 0

∴ The required equations of the tangents are y = 0, 9x + y – 27 = 0.

Question 24.

Show that the area of the triangle formed by the tangent at any point on the curve xy = c (c ≠ 0), with the coordinate axes is constant.

Solution:

Given, the equation of the curve is xy = c

Differentiating on both sides with respect to ‘x’

\(\frac{x}{2 x_1}+\frac{y}{2 y_1}\) = 1

∴ x-intercept, a = 2x_{1}, y-intercept, b = 2y_{1}

∴ The area of the triangle formed by this tangent and the coordinate axes = \(\frac{1}{2}\) |ab|

= \(\frac{1}{2}\) |2x_{1} . 2y_{1}|

= 2|x_{1}y_{1}|

= 2c (constant) (∵ P(x_{1}, y_{1}) lies on curve, x_{1}y_{1} = c)

Question 25.

Show that the equation of the tangent to the curve is \(\left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n\) = 2 (a ≠ 0, b ≠ 0) at the point (a, b) is \(\frac{\mathbf{x}}{\mathbf{a}}+\frac{\mathbf{y}}{\mathbf{b}}\) = 2. [Mar. ’18 (TS)]

Solution:

Given, equation of the curve is \(\left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n\) = 2

Differentiating on both sides with respect to ‘x’

∴ The equation of the tangent to the curve at the point (a, b) is y – y_{1} = m(x – x_{1})

y – b = \(\frac{-b}{a}\) (x – a)

ay – ab = -bx + ab

bx + ay = 2ab

\(\frac{b x}{a b}+\frac{a y}{a b}=2\)

\(\frac{x}{a}+\frac{y}{b}=2\)

Question 26.

Find the slope of the tangent to the curve y = \(\frac{x-1}{x-2}\) at x = 10.

Solution:

Given, equation of the curve is y = \(\frac{x-1}{x-2}\)

Differentiating on both sides with respect to ‘x’

Question 27.

Find the slope of the tangent to the curve y = x^{3} – x + 1 at the point whose x-coordinate is 2.

Solution:

Given, equation of curve is y = x^{3} – x + 1

Differentiating on both sides with respect to ‘x’

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\)(x^{3} – x + 1) = 3x^{2} – 1

Slope of tangent at x = 2 is m = \(\left(\frac{d y}{d x}\right)_{x=2}\)

= 3(2)^{2} – 1

= 12 – 1

= 11

Question 28.

Find the slope of the tangent to the curve y = x^{3} – 3x + 2 at the point whose x-coordinate is 3.

Solution:

Given, equation of curve is y = x^{3} – 3x + 2

Differentiating on both sides with respect to ‘x’

\(\frac{d y}{d x}=\frac{d}{d x}\left(x^3\right)-3 \frac{d}{d x}(x)+\frac{d}{d x}(2)\)

= 3x^{2} – 3 + 0

= 3x^{2} – 3

Slope of tangent at x = 3 is m = \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=3}\)

= 3(3)^{2} – 3

= 27 – 3

= 24

Question 29.

Find the slope of the normal to the curve x = a cos^{3}θ, y = a sin^{3}θ at θ = \(\frac{\pi}{4}\).

Solution:

Given, x = a cos^{3}θ

Differentiating on both sides with respect to ‘θ’

Question 30.

Find the point on the curve y = x^{3} – 11x + 5 at which the tangent is y = x – 11.

Solution:

Given, the equation of the curve is y = x^{3} – 11x + 5

Differentiating on both sides with respect to ‘x’

\(\frac{d y}{d x}=\frac{d}{d x}\left(x^3\right)-11 \frac{d}{d x}(x)+\frac{d}{d x}(5)\) = 3x^{2} – 11

The slope of the tangent at any point P(x, y) is m = \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{P}}\) = 3x^{2} – 11

Given the equation of the tangent is y = x – 11

Slope of the tangent is m = \(\frac{-\mathrm{a}}{\mathrm{b}}=\frac{-1}{-1}\) = 1

∴ 3x^{2} – 11 = 1

3x^{2} = 12

x^{2} = 4

x = 2

If x = 2, y = 2 – 11 = -9

∴ The required point is (2, -9).

Question 31.

Find the equations of tangent and normal to the curve y = x^{4} – 6x^{3} + 13x^{2} – 10x + 5 at the point (0, 5).

Solution:

Given, equation of the curve is y = x^{4} – 6x^{3} + 13x^{2} – 10x + 5 ……..(1)

Let the given point P(x, y) = (0, 5)

Differentiating (1) on both sides with respect to ‘x’

\(\frac{d y}{d x}=\frac{d}{d x}\left(x^4\right)-6 \frac{d}{d x}\left(x^3\right)+13 \frac{d}{d x}\left(x^2\right)\) – \(10 \frac{d}{d x}(x)+\frac{d}{d x}(5)\)

= 4x^{3} – 18x^{2} + 26x – 10

Slope of the tangent at P is m = \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{P}}\)

= 4(0)^{3} – 18(0)^{2} + 26(0) – 10

= -10

The equation of the tangent at P is y – y_{1} = m(x – x_{1})

y – 5 = -10(x – 0)

y – 5 = -10x

10x + y – 5 = 0

The equation of the normal at P is y – y_{1} = \(\frac{-1}{m}\) (x – x_{1})

y – 5 = \(\frac{-1}{-10}\) (x – 10)

10y – 50 = x

x – 10y + 50 = 0

Question 32.

Find the equations of tangent and normal to the curve x = cos t, y = sin t at t = \(\frac{\pi}{4}\).

Solution:

Given, x = cos t

Differentiating on both sides with respect to ‘t’.

\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = -sin t

Now, y = sin t

Differentiating on both sides with respect to ‘t’

\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = cos t

Question 33.

Find the equations of tangent and normal to the curve y = \(\frac{1}{1+x^2}\) at the point (0, 1).

Solution:

Given, equation of the curve is y = \(\frac{1}{1+x^2}\)

Differentiating on both sides with respect to ‘x’

The equation of the tangent at P is y – y_{1} = m(x – x_{1})

y – 1 = 0(x – 0)

y – 1 = 0

Equation of normal at P is y – y_{1} = \(\frac{-1}{m}\) (x – x_{1})

y – 1 = \(\frac{-1}{-0}\) (x – 0)

0(y – 1) = -x + 0

0 = -x

x = 0

Question 34.

Find the equations of tangent and normal to the curve xy = 10 at (2, 5). [Mar. ’17 (AP)]

Solution:

Given, the equation of the curve is xy = 10

y = \(\frac{10}{x}\) …….(1)

Let given point be P(x_{1}, y_{1}) = (2, 5)

Differentiating on both sides with respect to ‘x’

The equation of the tangent at P is y – y_{1} = m(x – x_{1})

y – 5 = \(\frac{-5}{2}\) (x – 2)

2y – 10 = -5x + 10

5x + 2y – 20 = 0

Equation of normal at P is y – y_{1} = \(\frac{-1}{m}\) (x – x_{1})

y – 5= \(\frac{-1}{\frac{-5}{2}}(x-2)\)

y – 5 = \(\frac{2}{5}\) (x – 2)

5y – 25 = 2x – 4

2x – 5y + 21 = 0

Question 35.

Find the equations of tangent and normal to the curve y = x^{3} + 4x^{2} at (-1, 3). [May ’15 (TS), ’14]

Solution:

Given, the equation of the curve is y = x^{3} + 4x^{2} …….(1)

Let the given point P(x_{1}, y_{1}) = (-1, 3)

Differentiating (1) on both sides with respect to ‘x’

\(\frac{d y}{d x}\) = 3x^{2} + 8x

Slope of tangent at P is m = \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{P}}\)

= 3(-1)^{2} + 8(-1)

= 3 – 8

= -5

The equation of the tangent at P is y – y_{1} = m(x – x_{1})

y – 3 = -5(x + 1)

y – 3 = -5x – 5

5x + y + 2 = 0

Equation of normal at P is y – y_{1} = \(\frac{-1}{m}\) (x – x_{1})

y – 3 = \(\frac{-1}{-5}\) (x + 1)

5y – 15 = x + 1

x – 5y + 16 = 0

Question 36.

If the slope of the tangent to the curve y = x log x at a point on it is \(\frac{3}{2}\), then find the equations of tangent and normal at the point.

Solution:

Given, the equation of the curve is y = x log x

Differentiating on both sides with respect to ‘x’

Question 37.

Show that the curves x^{2} + y^{2} = 2 and 3x^{2} + y^{2} = 4x have a common tangent at the point (1, 1).

Solution:

Given the equation of the first curve is x^{2} + y^{2} = 2

Differentiating on both sides with respect to ‘x’

2x + 2y . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0

x + y . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0

y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -x

\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{x}}{\mathrm{y}}\)

Slope of the tangent at P(1, 1) is m = \(\left(\frac{d y}{d x}\right)_P=\frac{-1}{1}\) = -1

The equation of the tangent at P is y – y_{1} = m(x – x_{1})

y – 1 = -1(x – 1)

y – 1 = -x + 1

x + y – 2 = 0

Given the equation of the second curve is 3x^{2} + y^{2} = 4x

Differentiating on both sides with respect to ‘x’

The equation of the tangent at P is y – y_{1} = m(x – x_{1})

y – 1 = -1(x – 1)

y – 1 = -x + 1

x + y – 2 = 0

∴ The curves x^{2} + y^{2} = 2 and 3x^{2} + y^{2} = 4x have a common tangent at the point (1, 1).

Question 38.

At a point (x_{1}, y_{1}) on the curve x^{3} + y^{3} = 3axy, show that the tangent is (\(x_1^2\) – ay_{1})x + (\(y_1^2\) – ax_{1})y = ax_{1}y_{1}. [Mar. ’17 (TS)]

Solution:

Given, equation of the curve is x^{3} + y^{3} = 3axy ……..(1)

Since P(x_{1}, y_{1}) lies on the curve (1). Then

Which is the required equation of tangent.

Question 39.

Show that the length of the subnormal at any point on the curve xy = a^{2} varies as the cube of the ordinate of the point.

Solution:

Given the equation of the curve is xy = a^{2}

Differentiating on both sides with respect to ‘x’

x . \(\frac{d y}{d x}\) + y . 1 = 0

x \(\frac{d y}{d x}\) = -y

\(\frac{d y}{d x} = \frac{-y}{x}\)

Let P(x, y) be any point on the curve.

The slope of the tangent at P is m = \(\left(\frac{d y}{d x}\right)_P=\frac{-y}{x}\)

Length of the sub-normal = |y_{1}m|

∴ Length of the sub-normal ∝ y^{3}. (Cube of ordinate)

Question 40.

Find the value of k so that the length of the subnormal at any point on the curve xy^{k} = a^{k+1} is a constant.

Solution:

Given the equation of the curve is xy^{k} = a^{k+1}

Differentiating on both sides with respect to ‘x’

∴ Length of the subnormal at any point is constant then k + 2 = 0

∴ k = -2.

Question 41.

Find the value of k, so that the length of the subnormal at any point on the curve y = a^{1-k} x^{k} is a constant.

Solution:

Given, the equation of the curve is y = a^{1-k} x^{k}

Differentiating on both sides with respect to ‘x’

The length of subnormal at any point is constant then 2k – 1 = 0

2k = 1

k = \(\frac{1}{2}\)

Question 42.

Find the lengths of sub tangent, and subnormal at a point ‘t’ on the curves x = a(cos t + t sin t), y = a(sin t – t cos t). [Mar. ’17 (TS); May ’15 (TS), ’14]

Solution:

Given that, x = a(cos t + t sin t)

Differentiating on both sides with respect to ‘t’

\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = a[(-sin t) + t(cos t) + sin t (1)]

= a[-sin t + t cos t + sin t]

= at cos t

y = a(sin t – t cos t)

Differentiating on both sides with respect to ‘t’

\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = a[cos t – (t(-sin t) + cos t (1)]

= a[cos t + t sin t – cos t]

= at sin t

\(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\not t \sin t}{\not t \cos t}=\tan t\)

Slope of tangent at any point t is m = \(\left(\frac{d y}{d x}\right)_t\) = tan t

Length of the sub tangent = \(\left|\frac{y_1}{m}\right|\)

= \(\left|\frac{\mathrm{a}(\sin \mathrm{t}-\mathrm{t} \cos \mathrm{t})}{\tan \mathrm{t}}\right|\)

= |a(sin t – t cos t) . cot t|

Length of the subnormal = |y_{1} . m| = |a (sin t – t cos t) . tan t|

Question 43.

Show that the condition for the orthogonality of the curves ax^{2} + by^{2} = 1 and a_{1}x^{2} + b_{1}y^{2} = 1 is \(\frac{1}{a}-\frac{1}{b}=\frac{1}{a_1}-\frac{1}{b_1}\). [Mar. ’19 (AP); Mar. ’16 (TS); May ’14]

Solution:

Given, curves are

ax^{2} + by^{2} = 1 ……..(1)

a_{1}x^{2} + b_{1}y^{2} = 1 ……….(2)

Let the curves intersect at P(x_{1}, y_{1}) then

\(\mathrm{ax}_1^2+\mathrm{by}_1^2=1, \mathrm{a}_1 \mathrm{x}_1{ }^2+\mathrm{b}_1 \mathrm{y}_1{ }^2=1\)

Differentiating (1) on both sides with respect to ‘x’

Question 44.

Find the angle between the curves x + y + 2 = 0 and x^{2} + y^{2} – 10y = 0. [Mar. ’14]

Solution:

Given equations of the curves are

x + y + 2 = 0 ………(1)

x^{2} + y^{2} – 10y = 0 ……….(2)

From (1), x = -y – 2

Substitute in (2)

(-y – 2) + y^{2} – 10y = 0

y^{2} + 4 + 4y + y^{2} – 10y = 0

2y^{2} – 6y + 4 = 0

y^{2} – 3y + 2 = 0

y^{2} – 2y – y + 2 = 0

y(y – 2) – 1(y – 2) = 0

(y – 2) (y – 1) = 0

y- 2 = 0 (or) y – 1 = 0

y = 2 (or) y = 1

If y = 2, x = -2 – 2 = -4

If y = 1, x = -1 – 2 = -3

Required points are P(-4, 2), Q(-3, 1)

Differentiating (1) on both sides with respect to ‘x’

1 + \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + 0 = 0

\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -1

Differentiating (2) on both sides with respect to ‘x’

Question 45.

Find the equation of tangent and normal to the curve y = \(\text { 2. } \mathrm{e}^{-\mathrm{x} / 3}\) at the point where the curve meets the y-axis. [Mar. ’16 (TS)]

Solution:

Equation of the curve is y = \(\text { 2. } \mathrm{e}^{-\mathrm{x} / 3}\)

The equation of the y-axis is x = 0

y = 2

Required point = (0, 2)

Also \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2}{3} \cdot \mathrm{e}^{\frac{-\mathrm{x}}{3}}\)

and slope m = \(\frac{-2}{3} \cdot \mathrm{e}^{\frac{-0}{3}}=\frac{-2}{3}\)

Equation tangent at p(0, 2) is y – y_{1} = m(x – x_{1})

y – 2 = \(\frac{-2}{3}\)(x – 0)

3y – 6 = 2x

2x + 3y – 6 = 0

Equation of the normal is y – y_{1} = \(\frac{-1}{m}\) (x – x_{1})

y – 2 = \(\frac{3}{2}\) (x – 0)

2y – 4 = 3x

3x – 2y + 4 = 0