TS Inter 2nd Year

TS Intermediate 2nd Year English Study Material Textbook Solutions Telangana

TS Inter 2nd Year English Textbook Answers

TS Intermediate 2nd Year English Textbook Solutions – Poetry

• Chapter 1 Goodbye Party for Miss Pushpa T S
• Chapter 2 One the Grasshopper and Cricket
• Chapter 3 Hiroshima Child
• Chapter 4 Awake
• Chapter 5 Fear

Inter 2nd Year English Textbook Answers Telangana – Prose

• Chapter 6 How to Avoid Foolish Opinions
• Chapter 7 The Awakening of Women
• Chapter 8 Solution to Plastic Pollution
• Chapter 9 The Religion of the Forest
• Chapter 10 Guilty

TS Inter 2nd Year English Short Stories

• Chapter 11 The Woman on Platform No 8
• Chapter 12 A Gift for Christmas
• Chapter 13 The Doctor’s Word
• Chapter 14 Lost
• Chapter 15 An Interview

TS Inter 2nd Year English Reading Comprehension

• Reading Comprehension Passages from Short Stories
• Reading Comprehension Unseen Passages
• Reading Understanding Advertisements
• Reading Understanding Non-verbal Data

TS Inter 2nd Year English Grammar with Answers

• Punctuation
• One-word Substitutes
• Idioms and Phrases
• Filling in Forms
• Curriculum Vitae
• Writing Descriptions
• Letter Writing
• Note Making
• Word Stress
• Dialogue Writing

TS Inter 2nd Year English Revision Tests I-V

• Revision Test-I
• Revision Test-II
• Revision Test-III
• Revision Test-IV
• Revision Test-V

TS Inter 2nd Year English Syllabus

Telangana TS Intermediate 2nd Year English Syllabus

Telangana State Board of Intermediate Education, Hyderabad
English-Intermediate 2nd Year
Syllabus (w.e.f. 2019-20)

TS Inter 2nd Year Study Material

TS Intermediate 2nd Year Accountancy Notes

• Chapter 1 Depreciation Notes
• Chapter 2 Consignment Accounts Notes
• Chapter 3 Accounting for Not-for-Profit Organisation Notes
• Chapter 4 Partnership Accounts Notes
• Chapter 5 Admission of a Partner Notes
• Chapter 6 Retirement and Death of a Partner Notes
• Chapter 7 Computerised Accounting System Notes

TS Intermediate 2nd Year Telugu Study Material Textbook Solutions Telangana

TS Inter 2nd Year Telugu Poems పద్య భాగం

• Poem 1 శ్రీకృష్ణ రాయబారం
• Poem 2 భగీరథ ప్రయత్నం
• Poem 3 జ్ఞానబోధ
• Poem 4 దుందుభి
• Poem 5 కోకిలా! ఓ కోకిలా !!
• Poem 6 ఆడపిల్లలంటేనే

TS Inter 2nd Year Telugu Textbook Lessons గద్య భాగం

• Chapter 1 మిత్రలాభం
• Chapter 2 తెలంగాణ సాహితీ వికాసం
• Chapter 3 నా సాహిత్య పరిశోధన
• Chapter 4 గోల్కొండ మధుర స్మృతులు
• Chapter 5 మా భాగోతంలో మేము
• Chapter 6 సృజనశీలత

TS Inter 2nd Year Telugu Non-Detailed ఉపవాచకం : యాత్రారచన

• Chapter 1 నా ప్రథమ విదేశయాత్ర

TS Intermediate 2nd Year Telugu Grammar వ్యాకరణాంశాలు

• అలంకారాలు
• ఛందస్సు
• భాషాభాగములు
• సంక్షిప్తీకరణ
• సంభాషణ రచనా నైపుణ్యం

TS Inter 2nd Year Telugu Syllabus

Telangana TS Intermediate 2nd Year Telugu Syllabus

Telangana State Board of Intermediate Education, Hyderabad
Telugu-Intermediate 2nd Year
Syllabus (w.e.f. 2019-20)

TS Inter 2nd Year Telugu Question Paper Design

ఇంటర్మీడియట్ – రెండవ సంవత్సరం
ద్వితీయ భాష – తెలుగు : భాగం-II
మాదిరి ప్రశ్నపత్రం – వివరణ

ఇంటరు రెండవ సంవత్సరము తెలుగు పేపరులో మొత్తం 16 ప్రశ్నలు ఉంటాయి. మూడు గంటల సమయంలో జవాబులు రాయాలి.
మొత్తం మార్కులు = 100

సూచనలు :
1. ప్రశ్నపత్రం ప్రకారం వరుసక్రమంలో సమాధానాలు రాయాలి.
2. ఒక్క మార్కు ప్రశ్నల జవాబులను కేటాయించిన ప్రశ్న కింద వరుసక్రమంలో రాయాలి.

I. పద్య భాగం నుండి రెండు పద్యాలు ఇస్తారు. అందులో ఒకదానికి ప్రతిపదార్థ తాత్పర్యాలను రాయాలి. (1 × 8 = 8 మార్కులు)

II. పద్య భాగం నుండి రెండు వ్యాసరూప సమాధాన ప్రశ్నలు ఇస్తారు. అందులో ఒక్క ప్రశ్నకు 20 పంక్తులలో సమాధానం రాయాలి. (1 × 6 = 6 మార్కులు)

III. గద్య భాగం నుండి రెండు వ్యాసరూప సమాధాన ప్రశ్నలు ఇస్తారు. అందులో ఒక్క ప్రశ్నకు 20 పంక్తులలో సమాధానం రాయాలి. (1 × 6 = 6 మార్కులు)

IV. ‘యాత్రారచన’ ఉపవాచకం నుండి నాలుగు ప్రశ్నలు ఇస్తారు. అందులో రెండు ప్రశ్నలకు 15 పంక్తులలో సమాధానాలు రాయాలి. (2 × 4 = 8 మార్కులు)

V. పద్య భాగం నుండి నాలుగు సందర్భ సహిత వ్యాఖ్యలు ఇస్తారు. అందులో రెండింటికి జవాబులు రాయాలి. (2 × 3 = 6 మార్కులు)

VI. ‘యాత్రారచన’ నుండి నాలుగు సందర్భసహిత వ్యాఖ్యలు ఇస్తారు. అందులో రెండింటికి సమాధానాలు రాయాలి. (2 × 3 = 6 మార్కులు)

VII. పద్య భాగంపై నాలుగు సంగ్రహ సమాధాన ప్రశ్నలు ఇస్తారు. అందులో రెండు ప్రశ్నలకు సంగ్రహంగా సమాధానాలు రాయాలి. (2 × 2 = 4 మార్కులు)

VIII. గద్య భాగంపై నాలుగు సంగ్రహ సమాధాన ప్రశ్నలు ఇస్తారు. అందులో రెండు ప్రశ్నలకు సంగ్రహంగా సమాధానాలు రాయాలి. (2 × 2 = 4 మార్కులు)

IX. పద్య భాగం నుండి ఒక వాక్యంలో సమాధానాలు రాయవలసిన ఎనిమిది ప్రశ్నలు ఇస్తారు. అందులో ఆరింటికి జవాబులు రాయాలి. (6 × 1 = 6 మార్కులు)

X. గద్య భాగం నుండి ఒక వాక్యంలో సమాధానాలు రాయవలసిన ఎనిమిది ప్రశ్నలు ఇస్తారు. అందులో ఆరింటికి జవాబులు రాయాలి. (6 × 1 = 6 మార్కులు)

XI. ఛందస్సు : మూడు పద్యములు ఇస్తారు. దానిలో ఒక పద్యానికి లక్షణాలు తెలిపి, ఉదాహరణతో సమన్వయించాలి. (1 × 6 = 6 మార్కులు)

XII. ఛందస్సుపై ఎనిమిది ఏకవాక్య సమాధాన ప్రశ్నలు ఇస్తారు. అందులో ఆరింటికి జవాబులు రాయాలి. (6 × 1 = 6 మార్కులు)

XIII. అలంకారములు : మూడు అలంకారాలు ఇస్తారు. అందులో ఒకదానికి లక్షణాలు తెలిపి ఉదాహరణతో సమన్వయించాలి. (1 × 6 = 6 మార్కులు)

XIV. అలంకారాలపై ఎనిమిది ఏకవాక్య సమాధాన ప్రశ్నలు ఇస్తారు. అందులో ఆరింటికి జవాబులు రాయాలి. (6 × 1 = 6 మార్కులు)

XV. సంక్షిప్తీకరణ : ఇచ్చిన విషయాన్ని 1/3 వంతుకు సంక్షిప్తం చేసి రాయాలి. (1 × 6 = 6 మార్కులు)

XVI. (అ) ఇచ్చిన పదాలు ఆధారంగా చేసుకుని సంభాషణ రాయాలి. (1 × 5 = 5 మార్కులు)
(ఆ) భాషాభాగాలపై ఐదు ఏకవాక్య సమాధాన ప్రశ్నలు ఇస్తారు. అన్నింటికీ జవాబులు రాయాలి. (1 × 8 = 8 మార్కులు)

మొత్తం = 100 మార్కులు

TS Inter 2nd Year Study Material

Students must practice these TS Inter 2nd Year Maths 2B Important Questions Chapter 7 Definite Integrals to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2B Definite Integrals Important Questions

Very Short Answer Type Questions

Question 1.
Evaluate \(\int_1^2 x^5\) dx
Solution:
\(\int_1^2 x^5 d x=\left[\frac{x^6}{6}\right]_1^2=\frac{2^6}{6}-\frac{1}{6}=\frac{64}{6}-\frac{1}{6}=\frac{63}{6}=\frac{21}{2}\)

Question 2.
Evaluate \(\int_0^\pi \) sinx dx
Solution:

Question 3.
Evaluate \(\int_0^a \frac{d x}{x^2+a^2}\)
Solution:

Question 4.
Evaluate \(\int_1^4 x \sqrt{x^2-1}\) dx
Solution:
Let x² – 1 – t ⇒ 2x dx dt then
Upper limit when x = 4 is t = 15.
Lower Limit when x = 1 is t = 0.

Question 5.
Evaluate \(\int_0^2 \sqrt{4-x^2}\) dx
Solution:
Let x= 2 sin θ = dx – 2cosθ dθ then

Question 6.
Show that \(\int_0^{\frac{\pi}{2}} \sin ^n x d x=\int_0^{\frac{\pi}{2}} \cos ^n x dx\)
Solution:

Question 7.
Evaluate \(\int_0^{\frac{\pi}{2}} \frac{\cos ^{\frac{5}{2}} x}{\sin ^{\frac{5}{2}} x+\cos ^{\frac{5}{2}} x}\) dx
Solution:

Question 8.
Evaluate \(\int_0^{\frac{\pi}{2}} \) x sin x dx
Solution:

Question 9.
Evaluate

Solution:

(iii) \(\int_0^{\frac{\pi}{2}} \sin ^6 x \cos ^4 x dx\)
Solution:

Question 10.
Find \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \cos ^4 x d x\)
Solution:

Short Answer Type Questions

Question 1.
Find \(\int_0^2\left(x^2+1\right) dx\) as the limit of a sum
Solution:

Question 2.
Evaluate \(\int_0^2 e^x dx\) as the limit of a sum.
Solution:

Question 3.
Lets define f : [0,1]→ R by
f(x) = 1 if x is rational
= 0 if x is irrational
then show that f is nor R Integrable over [0, 1].
Solution:
Let P = (x₀, x₁,…., x_n] be a partition of [0, 1].
Since between any two real numbers there exists rational and irrational numbers and
let t_i, s_i ∈ [X_i -i x_j] be the rational and irrational numbers.

Question 4.
Evalute \(\int_0^{16} \frac{x^{\frac{1}{4}}}{1+x^{\frac{1}{2}}}\) dx
Solution:
Let x = t⁴ then dx – 4t³ dt
Upper limit when x = 16 is t = 2.
and Lower limit when x = 0 is t = 0.

Question 5.
Evaluate \(\int_{-\frac{\pi}{2}}^\pi \sin\) |x| dx
Solution:
We have sin |x| = sin(-x) if x < 0
= sinx if x ≥ 0

Question 6.
Evaluate by using the method of finding definite integral as the limit of a sum.
\(\lim _{n \rightarrow \infty} \sum_{i=1}^n \frac{1}{n}\left(\frac{n-1}{n+1}\right)\)
Solution:

Question 7.
Evaluate \(\lim _{n \rightarrow \infty} \frac{2^k+4^k+6^k+\ldots+(2 n)^k}{n^{k+1}}\) using the method of finding definite integral as the limit of a sum.
Solution:

Question 8.
Evaluate \(\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right) \ldots\left(1+\frac{n}{n}\right)\right]^{\frac{1}{n}}\)
Solution:

Question 9.
Obtain Reduction formula for \(\int_0^{\frac{\pi}{2}} \sin ^n x d x\) and hence find

Solution:

Question 10.
Evaluate \(\int_0^a \sqrt{a^2-x^2} dx\)
Solution:
Let x = a sinθ then dx = a cosθ dθ
Upper limit when x = a is θ = \(\frac{\pi}{2}\)
and Lower limit when x = 0 is θ = 0

Question 11.
Find \(\int_{-a}^a x^2\left(a^2-x^2\right)^{3 / 2} dx\)
Solution:
Since f(x) = x² (a² – x²)^3/2 is an even function and f(- x) = f(x) we have
\(\int_{-a}^a x^2\left(a^2-x^2\right)^{3 / 2} d x=2 \int_0^a x^2\left(a^2-x^2\right)^{3 / 2} d x\)
Let x = a sin θ then dx = a cos θ dθ
∴ Upper limit when x = a is θ = \(\frac{\pi}{2}\)
Lower limit when x = 0 is θ = 0

Question 12.
Find \(\int_0^1 x^{3 / 2} \sqrt{1-x} dx\)
Solution:
Let x = sin²θ then dx = 2 sinθ cosθ dθ
Upper limit when x = 1 is θ = \(\frac{\pi}{2}\)
Lower Limit when x = θ is θ = 0.

Question 13.
Find the area under the curve f(x) = sin x in (0, 2π).
Solution:
Consider the graph of the function f(x) = sinx in [0, 2π];
we have sin x ≥ 0 ∀ x ∈ [0,π] and sin x≤0∀x∈[π,2π].

Question 14.
Find the area under the curve f(x) = cos x in [0, 2π].
Solution:

Question 15.
Find the bounded by the y = x² parabola the X- axis and the lines x = – 1, x = 2.
Solution:

Question 16.
Find the area cut off between the line y = 0 and the parabola y = x²– 4x + 3.
Solution:
The point of intersection of y – 0 and y = x² – 4x + 3 is given by x² – 4x + 3 = 0
= (x – 3)(x-1) = 0 = x = 1 or 3
y=x²– 4x + 3 ⇒ y+1 =  x²– 4x + 4 (x-2)²
Hence the equation represents a parabola
with vertex (2, -1) lies in IV quadrant.

Question 17.
Find the area bounded by the curves y = sin x and y = cos x between any two consecutive points of intersection.
Solution:
The given curves y = sin x and y = cosx and
tan x = 1 ⇒ x = \(\frac{\pi}{4}\)
∴ x = \(\frac{\pi}{4}\) and x = \(\frac{5 \pi}{4}\) are the two consecutive points of intersection.
Taking f(x) = sin x and g(x) cos x over \(\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right]\) we have
f(x)> g(x) ∀ x ∈\(\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right]\).
Hence the area bounded by y = sin x, y = cos x and the two points of intersection is

Question 18.
Find the area of one of the curvilinear rectangles bounded by y = sin x, y cos x and X-axis.
Solution:

Question 19.
Find the area of the right angled triangle with base b and altitude ‘h’ using the fundamental theorem of integral calculus.
Solution:

Question 20.
Find the area bounded between the curves y² – 1 = 2x and x = 0.
Solution:
The given curves are
y²-1-2x-2(x-0) ……………. (1)
= (y-0)² 2(x)+1=2 \(\left[\mathrm{x}+\frac{1}{2}\right]\)
(1) represents parabola with vertex \(\left(-\frac{1}{2}, 0\right)\)
Solving (1) and x = 0 we get
y² -1 = 0 ⇒ y = ±1
∴ The points of intersection are (0, 1), (0, -1).
The parabola meets the X- axis and y = 1 and y = – 1 and the curve is symmetric with respect to X – axis

Question 21.
Find the area enclosed by the curves y = 3x and y = 6x-x².
Solution:
Given curves are y³x and y=6x – x²
Solving 6x – x² = 3x = 3x – x² = 0
= x(3- x)=0 =x=0 or x=3
Taking f(x) = 3x and g(x) = 6x – x²
then g(x) ≥ 1(x) in [0, 3] and area enclosed between the line y = 3x and the parabola y = 6x-x² is

Long Answer Type Questions

Question 1.
Show that \(\int_0^{\frac{\pi}{2}} \frac{x}{\sin x+\cos x}\) dx =\(\frac{\pi}{2 \sqrt{2}} \log (\sqrt{2}+1)\)
Solution:

Question 2.
Evaluate \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\) dx
Solution:

Question 3.
Evaluate \(\int_{-a}^a\left(x^2+\sqrt{a^2-x^2}\right) dx\)
Solution:

Question 4.
Evaluate \(\int_0^\pi \frac{x \sin x}{1+\sin x}\) dx
Solution:

Question 5.
Find \(\int_0^\pi \mathbf{x}\) sin⁷ x cos ⁶ x dx.
Solution:

Question 6.
Find the area enclosed between y=x²-5x and y=4-2x.
Solution:
The graphs of curves are shown below.

Question 7.
Find the area bounded between the curves y = x², y = \(\sqrt{\mathbf{x}} \)
Solution:

Question 8.
Find the area bounded between the curves y²=4ax, x²= 4by(a>0,b>0).
Solution:

Students must practice these TS Inter 2nd Year Maths 2B Important Questions Chapter 6 Integration to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2B Integration Important Questions

Question 1.
Find ∫2x⁷ dx on R.
Solution:

Question 2.
Evaluate ∫cot²xdx on l⊂R {nπ:n∈Z)
Solution:
∫cot²xdx =∫(cosec²x – 1)dx
= ∫ cosec² x – ∫dx = – dx = – cotx – x + c

Question 3.
Evaluate \(\int\left(\frac{x^6-1}{1+x^2}\right)\) dx for x ∈ R
Solution:

Question 4.
Find ∫(1 – x)(4 -3x) (3 + 2x) dx ; x ∈R.
Solution:
(1- x)(4 – 3x)(3+2x) = 6x³ – 5x² – 13x + 12
∴ ∫ (1 – x)(4 – 3x)(3+2x)dx
=∫(6x³– 5x²_13x+ 12)dx

Question 5.
Evaluate \(\int\left(x+\frac{1}{x}\right)^3 d x, x>0\)
Solution:

Question 6.
Find \(\int \sqrt{1+\sin 2 x}\) dx on R.
Solution:

Question 7.
Find \(\int \frac{6 x}{3 x^2-2}\) dx on any interval I ⊂ R \(\left\{-\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}\right\}\)
Solution:

Question 8.
\(\int \frac{\left(\sin ^{-1} x\right)^2}{\sqrt{1-x^2}}\) dx on R
Solution:

Question 9.
Evaluate \(\int \frac{1}{1+(2 x+1)^2}\) dx on R.
Solution:

Question 10.
Evaluate \(\int \frac{x^5}{1+x^{12}}\) dx on R
Solution:

Question 11.
∫ cos³ sinx dx on R.
Solution:

Question 12.
Find \(\int\left(1-\frac{1}{x^2}\right) e^{\left(x+\frac{1}{x}\right)}\) dx on I where I = (0,∞)
Solution:

Question 13.
Evaluate \(\int \frac{1}{\sqrt{\sin ^{-1} x} \sqrt{1-x^2}}\)
Solution:

Question 14.
Evaluate \(\int \frac{\sin ^4 x}{\cos ^6 x} d x \), x ∈ I ⊂ R – {(2n+1) \(\frac{\pi}{2}\) …………… n∈Z}
Solution:

Question 15.
Evaluate ∫ sin² x dx on R.
Solution:

Question 16.
Find \(\int \frac{x^2}{\sqrt{x+5}}\) on (-5, ∞)
Solution:

Question 17.
Find  \(\int \frac{x}{\sqrt{1-x}}\) dx, x∈1=(0,1)
Solution:
Let 1 – x = t² over (0, 1)
then – dx = 2t dt and x = 1 – t²

Question 18.
Evaluate \(\int \frac{d x}{(x+5) \sqrt{x+4}}\) on (-4,∞)
Solution:
Let x + 4 – t² then dx – 2t dt
defined over (- 4, ∞)

Question 19.
Evaluate \(\int \frac{d x}{\sqrt{4-9 x^2}} \text { on } I=\left(-\frac{2}{3}, \frac{2}{3}\right)\)
Solution:

Question 20.
\(\int \frac{1}{a^2-x^2}\) dx for x E I = (- a, a).
Solution:

Question 21.
Evaluate \(\int \frac{1}{1+4 x^2}\) dx on R.
Solution:

Question 22.
Evaluate \(\int \sqrt{4 x^2+9}\) dx on R.
Solution:

Question 23.
Evaluate \(\int \frac{1}{\sqrt{4-x^2}}\) dx on (-2,2)
Solution:

Question 24.
Evaluate \(\int \sqrt{9 x^2-25} d x \text { on }\left(\frac{5}{3}, \infty\right)\)
Solution:

Question 25.
Evaluate \(\int \sqrt{16-25 x^2} d x \text { on }\left(-\frac{4}{5}, \frac{4}{5}\right)\)
Solution:

Question 26.
Find \(e^x \frac{(1+x)}{(2+x)^2}\) dx on l ⊂ R – {-2}
Solution:

Question 27.
Evaluate \(\int \frac{d x}{\sqrt{x^2+2 x+10}}\)
Solution:

Question 28.
Evaluate \(\int \frac{d x}{\sqrt{1+x-x^2}}\)
Solution:

Short Answer Type Questions

Question 1.
Evaluale \(\int\left(\frac{2 x^3-3 x+5}{2 x^2}\right)\) dx for x>0 and verify the result by differentiation.
Solution:

This is the given expression and the result is correct.

Question 2.
Evaluate \(\int \frac{1}{a \sin x+b \cos x}\) dx where a, b ∈ R and a² + b ² ≠ 0 on R
Solution:

Question 3.
Evaluate ∫ x sin^-1 x dx on (-1, 1).
Solution:
We use integration by parts by suitably
choosing y x and u = sin^-1 x so that

Question 4.
Evaluate ∫ x²cosx dx dx
solution:
Use integration by parts by choosing u x² and y = cos x, we get
∫ x² cos x dx dx = x² ∫cos x dx
\(-\int\left[\frac{d}{d x}\left(x^2\right) \int \cos x d x\right] dx\)
⇒ x² sin x – f 2x sin x dx
⇒ x² sinx-[2x(- cosx) – ∫2(- cosx)dx]
⇒ x² sinx+ 2xcosx – 2sinx+c
⇒ (x² -2) sinx + 2xcosx+ c
(again using integration by parts on ∫ 2x sin x dx)

Question 5.
Evaluate ∫ e^x sinx dx on R.
Solution:
Let I = ∫ e^x sinx dx. Then using integration by parts by taking u = e^x v = sin x we get

Question 6.
Find ∫ e^ax cos(bx +c) dx on R, where a,b,c are real numbers and b ≠ 0.
Solution:
Let I =∫ e^ax cos(bx +c) dx
using integration by parts by suitably choosing e^ax = u and cos (bx + c) = v, we get

Question 7.
Evaluate \(\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\) dx on (-1,1)
Solution:

Question 8.
Evaluate \(\int e^x\left(\frac{1-\sin x}{1-\cos x}\right)\) dx on I⊂R {2nπ : n ∈Z}.
Solution:

Question 9.
\(\int \frac{d x}{(x+5) \sqrt{x+4}}\)
Solution:

Question 10
Evaluate \(\int \frac{d x}{5+4 \cos x}\)
Solution:

Question 11.
\(\int \frac{d x}{3 \cos x+4 \sin x+6}\)
Solution:

Question 12.
Find \(\int \frac{d x}{d+e \tan x}\)
Solution:

Question 13.
Evaluate \(\int\left(\frac{\cos x+3 \sin x+7}{\cos x+\sin x+1}\right) d x\)
Solution:

Question 14.
Find \(\int \frac{x^3-2 x+3}{x^2+x-2}\) dx
Solution:
Integrand is a rational function in which the degree of the numerator Is greater than the denominator. Hence using synthetic division.

Question 15.
Find \(\int \frac{2 x^2-5 x+1}{x^2\left(x^2-1\right)}\) dx
Solution:

Question 16.
Find \(\int \frac{3 x-5}{x\left(x^2+2 x+4\right)}\) dx
Solution:

Long Answer Type Questions

Question 1.
Evaluate \(\int \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\) dx on l ⊂ R {-1,1}
Solution:

Question 2.
Find \(\int \frac{x^2 e^{m \sin ^{-1} x}}{\sqrt{1-x^2}}\) dx on (-1,1) where m is a real number.
Solution:

Question 3.
Evaluate \(\int \frac{x+1}{x^2+3 x+12}\) dx
Solution:

Question 4.
Evaluate \(\int(3 x-2) \sqrt{2 x^2-x+1}\) dx
Solution:

Question 5.
Evaluate \(\int \frac{2 x+5}{\sqrt{x^2-2 x+10}}\) dx
Solution:

Question 6.
Evaluate \(\int \frac{2 x+1}{x\left(x^2+4\right)^2}\) dx
Solution:

Question 7.
Find reduction formula for ∫ xⁿ e^ax dx, n being a positive integer and hence evaluate dx
Solution:

Question 8.
Obtain reduction formula for ∫ sinⁿ x dx for an integer n ≥ 2 and hence obtain ∫ sin⁴ xdx.
Solution:

Question 9.
Obtain reduction formula for ∫ sin^m x cosⁿ x dx for a positive integer m and integer n≥ 2.
Solution:

Question 10.
Obtain reduction formula for ∫ tanⁿ x dx for an integar n ≥ 2 and hence find ∫ tan⁶ x dx.
Solution:

Question 11.
Obtain reduction formula for ∫ secⁿ x dx for n ≥ 2 and hence evaluate ∫ sec⁵ xdx
Solution:

Students must practice these TS Inter 2nd Year Maths 2B Important Questions Chapter 5 Hyperbola to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2B Hyperbola Important Questions

Short Answer Type Questions

Question 1.
If e, e₁ are the eccentricities of a hyperbola and its conjugate hyperbola, prove that \(\frac{1}{e^2}+\frac{1}{e_1^2}=1\)
Solution:
Let e, e₁ be the eccentricities of hyperbola

Question 2.
Find the equations of the tangents to the hyperbola 3x² – 4y² = 12 which are (I) parallel and (H) perpendicular to the line y = x – 7.
Solution:
Equation of given hyperbola is \(\frac{x^2}{4}-\frac{y^2}{3}=1\)
So that a² = 4, b² = 3 and equation to the given line y = x  – 7 and slope is ‘1’.

(i) Slope of the tangents which are parallel to the given line is ‘1’.
∴ Equation of tangents are
y = mx ± \(\sqrt{a^2 m^2-b^2}\)
⇒ y=x± \(\sqrt{4-3}\) and
⇒ y = x ± 1

(ii) Slope of the tangent which are perpendicular to the given line is – 1.
∴ Equations of tangents which are perpendicular to the given line are
y = (-1) x ± \(\sqrt{4(-1)^2-3}\)
x+y = ±1

Question 3.
A circle on the rectangular hyperbola xy = 1. In the points (Xr Yr)’ (r = 1, 2, 3, 4) Prove that x₁ x₂ x₃ x₄ = y₁ y₂ y₃ y₄ = 1.
Solution:
Let the circle be x² + y² = a².
Since \((\mathrm{t}, \frac{1}{t})\)(t ≠ 0) lies on xy= 1, the points of intersection of the circle and the hyperbola are given by
\(t^2+\frac{1}{t^2}=a^2\)

Long Answer Type Questions

Question 1.
Find the centre, eccentricity, foci, directrices and the length of latus rectum of the following hyperbolas.
4x² – 9y² – 8x -32 = 0
Solution:
Given equation is 4x² – 9y² –  8x – 32 = 0
⇒ 4x² – 8x -9y² = 32
⇒ 4(x -2x)-9y=32
⇒ 4(x²-2x+ 1) – 9y² = 32 + 4 = 36
⇒ 4 (x-1)2 – 9y² = 36
\(\frac{(x-1)^2}{9}-\frac{(y-0)^2}{4}=1\)
∴ Centre of the hyperbola = (1, 0)
The semi-transverse axis a = 3, and the semiconjugate axis b = 2.
∴ \(e=\sqrt{\frac{a^2+b^2}{a^2}}=\sqrt{\frac{9+4}{9}}=\sqrt{\frac{13}{3}}\)
Coordinates of foci (h ± ae, k)

(ii) 4(y+3)² -9(x-2)²= 1
Solution:
The equation can be written as

Question 2.
(i) If t be line lx+ my+n= 0 is a tangent to the hyperbola \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) then show that
a²l² – b²m² = n²

(ii) If the lx + my = t is a normal to the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) then show that
\(\frac{a^2}{l^2}-\frac{b^2}{m^2}=\left(a^2+b^2\right)^2\)
Solution:
(i) Let the line lx + my + n = 0 ……………….. (1) is a tangent to the hyperbola S = \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) at P(θ).
Then the equation of tangent at P(θ) is
\(\frac{x}{a}\) secθ – \(\frac{x}{b}\) tanθ – 1 = θ ………….. (2)
Since (1) and (2) represent the same line,

Question 3.
Prove that the point of intersection of two per perpendicular tangents to the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) lies on the circle x +y =a – b.
Solution:
Let P (x₁, y₁) be a point of intersection of two perpendicular tangents to the hyperbola
The equation of any tangent to S = 0 is of the form
\(S \equiv \frac{x^2}{a^2}-\frac{y^2}{b^2}-1=0\)
The equation of any tangent to S = 0 is of the form

This is a quadratic equation ¡n ‘m’ which has two roots m₁, m₂ (say) which corresponds to slopes of tangents.

Question 4.
If your points be taken on a rectangular hyperbola such that the chords joining any two points is perpendicular to the chord Joining the other two, and if α , β , γ and δ be the inclinations to either asymptote of the straight lines joining these points to the centre, prove that tanα, tanβ, tanγ , tanδ = 1
Solution:
Let the equation of rectangular hyperbola be x² – y² = a². By rotating the X – axis and Y – axis about the orgin through an angle \(\frac{\pi}{4}\) in the clockwise direction the equation x² – y² = a² will be transformed to xy = C².

Since \(\overline{\mathrm{AB}}\) is perpendicular \(\overline{\mathrm{CD}}\) we have
\(\left(-\frac{1}{t_1 t_2}\right)\left(-\frac{1}{t_3 t_4}\right)=-1\)
⇒ t₁ t₁ t₁ t₁ = – 1 ………………. (1)
We have the coordinate axis as the a asymptotes of the curves.
If \(\overline{\mathrm{OA}}, \overline{\mathrm{OB}}, \overline{\mathrm{OC}}, \overline{\mathrm{OD}}\) make angles α, β, γ and δ with positive direction of X-axis then tanα, tanβ, tanγ, and tanδ are the slopes.

If \(\overline{\mathrm{OA}}, \overline{\mathrm{OB}}, \overline{\mathrm{OC}}, \overline{\mathrm{OD}}\) make angles α, β, γ and δ with the other asymptote the Y – axis then cot α, cot β, cot γ, cot δ are the respective slopes.
So that cot α cot β cot γ cot δ = tan α tan β tan γ tan δ = I

Students must practice these TS Inter 2nd Year Maths 2B Important Questions Chapter 4 Ellipse to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2B Ellipse Important Questions

Very Short Answer Type Questions

Question 1.
If the length of the latus rectum is equal to half of its minor axis of an ellipse in the standard form, then find the eccentricity of the ellipse.
Solution:
Let \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) + 1(a > b) be the ellipse in its standard form.
Given that the length of the latus rectum = \(\frac{1}{2}\) (minor axis)

Question 2.
The orbit of the Earth is an ellipse with eccentricity with the \(\frac{1}{60}\) Sun at one of its foci, the major axis being approximately 186 x 10⁶ miles in length. Find the shortest and longest distance of the Earth from the Sun.
Solution:
Let the earths orbit be an ellipse given by
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) + 1(a > b)
Since the major axis is 186 x 10⁶ miles
we have 2a = 186 x 10⁶ ⇒ a = 93 x 10⁶ miles
If e is the eccentricity of ellipse then e = \(\frac{1}{60}\)
The longest and shortest distances of the earth from the sun are respectively a + ae and a – ae.
Here the longest distance of earth from the sun = a +ae \(\left(1+\frac{1}{60}\right)\)
= 9445 x 10⁴ miles and shortest distance of earth from the sun = a – ae
= 93 x 10⁴  \(\left(1-\frac{1}{60}\right)\)
= 9145 x 10⁴  miles

Short Answer Type Questions

Question 1.
Find the eccentricity, coordinates of foci, length of latus rectum and equations of directrices of the following ellipse
9x² + 16y²– 36x + 32y – 92 = 0
Solution:
Given equation of ellipse is
9x² + 16y²– 36x + 32y – 92 = 0
which can be written as

(ii) 3x+ y² – 6x -2y -5 = 0
Solution:
Given equation can be written as
3x² – 6x + y² – 2y = 5
⇒ 3(x²– 2x) + (y²– 2y) = 5
⇒ 3(x² -2x + 1)+(y²-2y+1) = 5 + 4
⇒ 3(x – 1)² + (y – 1)² = 9
⇒ \(\frac{(x-1)^2}{3}+\frac{(y-1)^2}{9}=1\)
which is of the form
\(\frac{(\mathrm{x}-\mathrm{h})^2}{\mathrm{a}^2}+\frac{(\mathrm{y}-\mathrm{k})^2}{\mathrm{~b}^2}=1\)
which a² = 3 and b² = 9 and a < b.
Also (h, k) = (1, 1); eccentrIcity

Question 2.
Find the equation of the ellipse referred to its major and minor axes as the coordinate axes X, Y – respectively with latus rectum of length 4 and distance between foci \(4 \sqrt{2}\).
Solution:
Let the equation of ellipse be

Question 3.
C is the centre, A A’ and B B’ are major and minor axis of ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\).If
PN is the ordinate of a point P on the ellipse then show that \(\frac{(\mathrm{PN})^2}{(\mathrm{AN})(\mathrm{AN})}+\frac{(\mathrm{BC})^2}{(\mathrm{CA})^2} \)
Solution:
Let P(θ) = (a cos θ, b sin θ) be any point on the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

Question 4.
S and T are the foci of an ellipse and B is one end of the minor axis. If STB is an equilateral triangle, then find the
eccentricity of the ellipse.
Solution:
Let \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 :(a>b) be an ellipse whose foci are S and T. B is the end of minor axis such that STB is an equilateral triangle.
than SB = ST = SB. Also S = (ae, 0).
T = (- ae. 0) and B = (0, b).

Question 5.
Show that among the points on the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\)=1,(a>b) ;(-a, 0) is the farthest point and (a, 0) is the nearest point form the focus (ae, 0).
Solution:
Let P(x, y) be any point on the ellipse so that < x < a and S = (ae, 0) is the focus. Since (x. y) is on the ellipse.

Maximum value of SP is a + ae when P(-a.0)
and Minimum value of SP is a – ae when P (a. 0).
The nearest point is (a, 0) and the farthest point is (-a, 0).

Question 6.
Find the equation of tangent and normal to the ellipse 9x² + 16y² = 144 at the end of latus rectum in the first quadrant.
Solution:
Given equation of ellipse is 9x² + 16y² = 144

Question 7.
If a tangent to the ellipse = \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)\) meets its major axis and minor axis at M and N respectively then prove that \(\frac{a^2}{(\mathrm{CM})^2}+\frac{b^2}{(\mathrm{CN})^2}=1\) where C is the centre of the ellipse.
Solution:

Question 8.
Find the condition for the line,
(i) lx + my + n = 0 to be a tangent to the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
(ii) lx + my + n = 0 to be a normal to the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
Solution:
Let lx + my + n = 0 be a tangent at
P (θ) (a cos θ . b sin θ) on the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

(ii) Let lx + myn = 0 be a normal to the ………………  (2)
Ellipse at the point P (θ). Then equation of normal at ‘θ’ is

Question 9.
If PN is the ordinate of a point P on the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) and the tangent at P
meets the x-axis at T then show that (CN) (CT) = a² where C is the centre of ellipse.
Solution:
Let P (θ) = P (a cos θ, b sin θ) be a point on the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) .
Then the equation of the tangent at p (θ) is.

Question 10.
Show that the points of intersection of the perpendicular tangents to any ellipse lie on the circle.
Solution:
Let the equation of ellispe be \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) (a>b)
Any tangent to the above ellipse is of the form y = mx ± \(\pm \sqrt{a^2 m^2+b^2}\)
Let the perpendicular tangents intersect at

This being a quadratic is ‘rn has two roots m₁ and m₂ which corresponds to the slopes of tangents drawn from P to ellipse then

(∵ Product of slopes = – 1 for perpendicular tangents)
⇒ x₁²+y₁² = a² +b²
∴ Locus of (x₁, y₁) is x² + y² = a²+ b² which is a circle.

Long Answer Type Questions

Question 1.
If θ₁, θ₂ are the eccentric angles of the extremeties of a focal chord (other than the verticles) of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) (a> b) and e is its eccentricity. Then show that

Solution:

Let P(θ₁), Q(θ₂) be the two extremeties of a focal chord of the ellipse
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)\)
∴ P = (acos θ₁, b sin θ₁),(θ₁ ≠ 0)
Q = (a cos θ₂, b sin θ₂), (θ₂ ≠ π)
and focus S = (ae, 0). Now PQ is a focal chord and hence P, S. Q are collinear.
∴ Slope of \(\overline{\mathrm{PS}}\) = slope of \(\overline{\mathrm{SQ}}\)
\(\frac{b \sin \theta_1}{a\left(\cos \theta_1-\mathrm{e}\right)}=\frac{\mathrm{b} \sin \theta_2}{\mathrm{a}\left(\cos \theta_2-\mathrm{e}\right)}\)

Question 2.
If the normal at one end of a latus rectum of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) passes through one end of the minor axis, then show that e⁴ + e² = 1 (e is the eccentricity of the ellipse)
Solution:
Let L be the one end of the latus rectum of \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). Then the coordinates of
\(L=\left(a e, \frac{b^2}{a}\right)\)
∴Equation of normal at L is

Question 3.
If a circle is concentric with the ellipse, find the inclination of their common tangent to the major axis of the ellipse.
Solution:
Let the circle x² + y² = r² and the ellipse be \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with a> b.
The major axis of ellipse is X – axis.
If r< b <a, then the circle lies completely in the ellipse making no common tangents.
If b < a < r (ellipse lies completely in circle) no common tangent is passive.

Case (i) : If b <r <a

Let one of the common tangent make angle θ with positive X- axis and suppose the equation of tangent to the circle be x Cos α + y sin α = r where a is the angle made by the radius of circle with positive X – axis.
∴ \(\theta=\frac{\pi}{2}+\alpha \text { (or) } \theta=\alpha-\frac{\pi}{2}\)
Since x cos α + y sina r touches the ellipse also, we have a² cos²a + b² sin² = r²

Case (ii): When r = a the circle touches the ellipse at the ends of major axis of the ellipse so that the common tangents are x = ± a and θ = \(\frac{\pi}{2}\)

Case (iii): When r = b, the circle touches the ellipse at the ends of minor axis of ellipse so that common tangents
y = ± b making θ = 0.

Students must practice these TS Inter 2nd Year Maths 2B Important Questions Chapter 3 Parabola to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2B Parabola Important Questions

Very Short Answer Type Questions

Question 1.
Find the coordinates of the vertex and focus and the equations of the directrix and axis of the following parabolas.
(i) y² = 16x
Comparing with y² = 4ax, we have 4a = 16
a = 4
Vertex = (0, 0)
Focus = (a, 0) = (4, 0)
Equation of clirectrix is x + a = 0 ⇒ x + 4 = 0
Equation of axis is y = 0.

(ii) x² =-4y
Comparing with x² = – 4ay, we have 4a = 4 ⇒ a = I
∴ Vertex = (0. 0)
Focus =(0,-a)=(0,-1)
Equation of directrix is y-a=0=y-1 =0
Equation of axis is x – 0.

(iii) 3x- 9x + 5y – 2 = 0
Given equation is 3x² – 9x – 5y +2
= 3(x² – 3x) =- \(5\left(\mathrm{y}-\frac{2}{5}\right)\)

(i) Vertex (h, k) \(\left(\frac{3}{2}, \frac{7}{4}\right)\)
(ii) Focus=(h,k,-a) = \(\left(\frac{3}{2}, \frac{7}{4}-\frac{5}{12}\right)=\left(\frac{3}{2}, \frac{4}{3}\right)\)
(iii) Equation of directrix is y – k = 0
⇒ 6y – 13= 0
(iv) Equation of axis x – h = 0
= x = \(\frac{3}{2}\) – 0 ⇒ 2x – 3 = 0

(iv) y² – x + 4y + 5 = 0
The given equation can be written as
y² + 4y – x – 5
9 + 4y + 4 – x – 1
(y+2)² – (x-1)
Comparing with (y – k)² – 4a (x – h)
we have h=1, k = -2, a= \(\frac{1}{4}\)
(i) Vertex = (h, k) – (1, -2)
(ii) Focus = (h+a,k) = \(\left(\frac{5}{4},-2\right)\)
(iii) Directrix is x – h + a = 0 ⇒ 4x – 3 = 0
(Iv) Equation of axis is y-k=0 ⇒ y+2 = 0

Question 2.
Find the equation of parabola whose vertex is (3-2) and focus is (3,1)
Solution:
The abscissae of vertex and focus is ‘3’.
Hence the axis of the parabola is x -3, a line parallel to y-axis, focus is above the vertex.
a = SA = \(\sqrt{(3-3)^2+(1+2)^2}=3\)
(∵ A=(3,-2) S= (3,3))
∴ Equation of the parabola
(x-h)² =4a (y-k)
(x-3)²– 12(y+2)
(∵ A = (h,k) =3,-2 and S (h, k-a)]

Question 3.
Find the coordinates of the points on the parabola y² = 2x whose focal distance is \(\frac{5}{2}\)
Solution:
Let P(x₁, y₁) be any point on the parabola
y² = 2x whose focal distance is \(\frac{5}{2}\) then
\(y_1^2=2 x_1\) and \( x_1+a=\frac{5}{2} \Rightarrow \frac{5}{2}-\frac{1}{2}\)
Where 4a=2 ⇒ a = \(\frac{1}{2}\)
∴ x₁= 2, and since y_1² = 2x₁ we have
y_1² = 4 = y₁ = ± 2
∴ The required points are (2, 2), (2, -2).

Short Answer Type Questions

Question 1.
Find the equation of the parabola passing through the points (- 1, 2), (1, – 1) and (2, 1) and having its axis parallel to the x-axis.
Solution:
Since the axis is parallel to x – axis the equation of the parabola is in the form of
x – ly²+my+n ……………. (1)
Given that the parabola passes through the points (-1, 2), (1, -1) and (2, 1) we have
– 1 = 14l+2m+n ……………………. (2)
1= l – m+n ……………….. (3)
2=l+m+n …………………….. (4)
From (2) and (3), -2 = 3l + 3m

∴  Equation of required parabola is
\(x=-\frac{7}{6} y^2+\frac{1}{2} y+\frac{8}{3}\)

Question 2.
A double ordinate of the curve y² = 4ax is of length 8a. Prove that the lines from the vertex to its ends are at right angles.
Solution:

Question 3.
Find the condition for the straight line lx + my + n=0 to be a tangent to the parabola y² = 4ax and find the coordinates
of the point of contact.
Solution:
Let the line ix + my + n = 0 …………….. (1) touches
y² = 4ax …………….. (2) at the point (x₁, y₁).
Then the equation of tangent at (x₁, y₁) to the parabola
y²=4ax is yy₁– 2ax – 2ax₁=0 …………. (2)
Since (1) and (2) represent the same line, coefficients are proportional.

Question 4.
Show that the straight line 7x+6y=13 is a tangent to the parabola y²– 7x – 8y + 14 = 0 and find the point of contact
Solution:
Given line is 7x + 6y – 13 = 0 ……………….. (1)
and parabola is y² – 7x – 8y+ 14 = 0 ………………..(2)
From (1),

Point of contact is (1, 1) and this point satisfies (1) and (2). Hence (1) is tangent to the parabola (2).

Question 5.
From an external point P, tangents are drawn to the parabola y² = 4ax and these tangents make angles θ₁, θ₂ with its axis such that tan θ₁+ tan θ₂ is a constant b. Then show that P lies on the line y = bx.
Solution:
Let P(x₁, y₁) be the external point.
Any tangent to the parabola is of the form
y=mx+ \(\frac{\mathrm{a}}{\mathrm{m}}\) =mx₁– my₁+a = 0
Let the roots of (1) be 4 m₁, m₂ ………………  (1)
Then m₁ + m₂ = \(\frac{y_1}{x_1}\) and m₁ + m₂ = \(\frac{\mathrm{a}}{\mathrm{x}_1}\)
The tangents make angles θ₁ and θ₂ with the axis such that m₁ = tan θ₁ and m₂= tan θ₂
Given tan θ₁ + tan θ₂ = b
⇒ m₁ + m₂ = b
\(\frac{\mathrm{y}_1}{\mathrm{x}_1}\) = b ⇒ y₁ = bx₁
∴ Locus of (x₁, y₂) is y = bx.

Question 6.
Show that the common tangents to the parabola y² = 4ax and x² = 4by is xa^1/3 + yb^1/3 + xa^2/3 yb^2/3
Solution:
The equations of the parabolas are
y² = 4ax ………………….. (1)
and x² = 4by ………………… (2)
Equation of any tangent to (1) is of the form
y=mx+\(\frac{\mathrm{a}}{\mathrm{m}}\) ………….. (3)
lf the line (3) is tangent to (2) also the point of intersection of (2) and (3) coincide.
From (3) and (2)
x² = 4b \(\left(\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\right)\)
⇒ mx² – 4bm²x – 4ab = 0
which should have equal roots.
∴ Therefore the discriminant is zero.

Long Answer Type Questions

Question 1.
Find the equation of a parabola in his standard form.
Solution:

Let S be the focus, l be the directrix as shown in the figure. Let Z be the projection of S on I and A be the mid point of \(\overline{\mathrm{SZ}}\). A lies on the parabola since SA = AZ. A is the vertex of the parabola. YY’ is parallel to the clirectrix ‘l’.
Let S=(a,0), (a>0) then Z =(-a, 0) and the equation of the directrix is x + a = 0.
Let P(x, y) be a point on the parabola and PM is the perpendicular distance from P to x + a = 0.
Then by definition \(\frac{\mathrm{SP}}{\mathrm{PM}}\) = e = 1
= SP² = PM²
= (x- a)² +(y-0)² = (x+a)²
= y² =(x+a)²-(x-a)² = 4ax
Conversely also if P(x, y) is a point such that y² = 4ax then SP = PM.
∴ P(x, y) lies on the locus. Equation of parabola in standard form is y² = 4ax.

Question 2.
(i) If the coordinates of the ends of a focal chord of the parabola y² = 4ax are (x₁, y₁) and (x₂, y₂) then prove that
x₁x₂ = a² and y₁y₂ = – 4a².
(ii) For a focal chord PQ of the parabola y² = 4ax if SP =l and SQ = l’ then prove that \(\frac{1}{l}+\frac{1}{l}=\frac{1}{a}\)
Solution:
(i) Let P(x₁, y₁) = (at₁², 2at₁)
and Q(x₂, y₂) = (at₂², 2at₂)
be two end points of a focal chord and S(a, 0) be the focus of the parabola y² = 4ax. PSQ is a focal chord P, S, Q are collinear.

(ii) Let P(at²₁, 2at₁) and Q (at²₂, 2at₂) be the extremities of a focal chord of parabola such that t₁,t₂ = – 1.

Question 3.
If Q is the foot of the perpendicular from a points P on the parabola y² = 8(x -3) to its directrix. S is the focus of parabola and if SPQ is an equilateral triangle then find the length of side of triangle.
Solution:
Given equation of parabola is
(y – 0)² = 8(x – 3)
which is of the form (y – k)² = 4a (x – h)
where 4a = 8 ⇒ a = 2
∴ Vertex = (h, k) = (3,0)
and focus = (h + a, k) = (3 +2, 0) =  (5, 0)

Since PQS is an equilateral triangle.
\(\angle \mathrm{SQP}=60^{\circ} \Rightarrow \angle \mathrm{SQZ}=30^{\circ}\)
Also in ΔSZQ , we have sin 30° = \(\frac{\mathrm{SZ}}{\mathrm{SQ}}\)

Hence length of each side of the triangle is 8

Question 4.
The cable of a uniformly loaded suspension bridge hangs In the form of a parabola. The roadway which is horizontal and 72mt. long is supported by vertical wires attached to the cable. The longest being 30 mts. and the shortest being 6 mts. Find the length of the supporting wire attached to the roadway 18 mts. from the middle.
Solution:

Let AOB be the cable [O is the lowest point and A, B are highest points). Let PRQ be the suspension bridge suspended with PR – RQ = 36 mts.
PA = QB = 30 mts (longest wire)
OR = 6 mts (shortest wire)
∴ PR = RQ = 36 mts. We take O as origin of coordinates at O, X-axis along the tangent \(\overline{\mathrm{RO}}\). So the equation of the cash is x² = 4ay for some a> 0.
∴ B(36,24) and B is a point on x²=4ay
We have (36)² = 4a(24)
⇒ 4a = \(\frac{36 \times 36}{24}\) = 54 mts
If RS = 18 mts and SC is the vertical through S meeting the cable at ‘C’ and the X- axis at D.
Then SC is the length of the required wire.
Let SC = l mts then DC = L – 6 mts.
∴ C = (18, 1 – 6) which lies on x² = 4ay
⇒ (18)² = 4a (l – 6)
⇒ (18)²= 54 (l -6)
⇒ l – 6 = \(\frac{18 \times 18}{54}\) = 6
⇒ l = 12.

Question 5.
Prove that the normal chord at the point other than origin whose ordinate is equal to Its abscissa subtends a right angle at the focus
Solution:
Let the equation of the parabola
y² = 4ax ……………. (1)
and P(at², 2at) be any point
Ordinate – abscissa
= 2at = at² = t=0 or t = 2
But t ≠ 0, Hence the point (4a, 4a) at which normal is
⇒  y + 2x = 2a(2) + a(2)³
⇒  y + 2x = 4a + 8a
⇒  y=12a -2x …………….. (2)
∴ From (1) we get
(12a – 2x)² – 4ax
⇒ 4x² – 48ax + 144a² = 4ax
⇒ 4x² – 52ax+ 144a²= 0
⇒ x² – 13ax + 36a² = 0
⇒ (x – 4a) (x – 9a) – 0
⇒ x = 4a, 9a
Correspondingly we get y = 12a – 8a = 4a and y = 12a – 18a= – 6a
Hence the other points of intersection of normal at P(4a, 4a) is Q(9a – 6a).
We have S = (a, 0)

Question 6.
Prove that the area of the triangle formed by the tangents at (x₁, y₁), (x₂, y₂) and (x₃,y₃) to the parabola y² = 4ax
\(\frac{1}{16 a}\left|\left(y_1-y_2\right)\left(y_2-y_3\right)\left(y_3-y_1\right)\right|\)  sq. unit
Solution:
Let A (x₁, y₁) = (at₂², 2at₁)
B (x₂, y₂) =  (at₂², 2at₂)
C (x₃, y₃) – (at₃², 2at₃) be the three points on the parabola y²=4ax.
The equations of tangents at A, B, C are

∴ Point of intersection of tangents at A and B is P(at₁t₂, a(t₁ + t₂)).
Similarly the points of intersection of tangents at B and C as well as A & C are

Question 7.
Prove that the two parabolas y² = 4ax and x² = 4by intersect other than the origin) at tan^-1
\(\left(\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right)\)
Solution:
Take a>0, b>0
and y²=4ax ………………….. (1)
and x² = 4by are ……………. (2)
the given parabolas.
Solving (1) and (2) we get the point of intersection other than the origin.

Question 8.
Prove that the orthocentre of the triangle formed by any three tangents to a parabola lies on the directrix of the parabola.
Solution:
Let y² = 4ax be the parabota and
A = (at₁², 2at₁), B = (at₂², 2at₂),
C = (at₃², 2at₃) be any three points on It.
If P, Q, R are the points of intersection of tangents at A and B, B and C, C and A then
P = [at₁t₂, a(t₁ + t₂)] Q = [at₂t₃, a(t₂ + t₃)] R = [at₁t₃, a(t₁ + t₃)]
Consider the ΔPQR then equation \(\overline{\mathrm{QR}}\) of (Tangent at C) is
x – yt₃ + at₃² = 0
∴ Altitude through P of Δ PQR is
t₃x + y = at₁t₂t₃ + a(t₁ + t₂) …………….. (1)
∴ Slope = \(\frac{1}{t_3}\) and equation is
y – a(t₁+ t₂) = – t₃ [x – at₁t₂]
y+xt₃ – at₁t₂t₃+a(t₁ +t₂)
Similarly the altitude through Q is
t₁x + y – at₁t₂t₃ + a(t₂+ t) …………… (2)
Solving (1) and (2)
x(t₃ – t₁) = a(t₁– t)
⇒ x = – a
Hence the orthocentre of ΔPQR with x coordinate as – ‘a’ lies on the directrix of the parabola.

Students must practice these TS Inter 2nd Year Maths 2B Important Questions Chapter 2 System of Circles to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2B System of Circles Important Questions

Very Short Answer Type Question

Question 1.
Find the angle between the circles x²+y²+4x-14y+28=0
x²+y²+4x – 5 = 0
Solution:
Comparing with general equation
x²+y²+2gx+2fy+c=0, we have
g=2, 1=-7, c=28, C₁=(-2,7)
g’=2, f’=0, c=-5, C₂=(-2,0)
Let θ be the angle between circles (1) and (2) then

Question 2.
If the angle between the circles x²+y²-12x- 6y+41=0 ………………..(1) x²+y²+kx+6y-59=0 …………… (2) is 45° find k.
Solution:


⇒ k² + 27². 18k²
⇒ 17k² = 27²
⇒ k²=16
k = ± 4

Short Answer Type Questions

Question 1.
find the equation of the circle passing through the points of the intersection of the circles
x²+y²– 8x-6y+21=0 …………… (1)
x²+y²– 2x-15=0 …………. (2) and (1, 2)
Solution:
The equation of circle passing through the point of intersection o! circles S = 0, S’ = 0 is S+ λS’+ 0 where λ is a parameter.
∴(x²+y²-8x-6y+21) +λ(x²+y²-2x-15)=0 ……………….. (1)
If this passes through (1, 2) then
(1+4-8-12+21)+λ(1+4-2-15)=0
= 6-12λ=0 ⇒ λ= \(\frac{1}{2}\)
Hence the equation of the required circle
is (x²+y²-8x-6y+ 21) + \(\frac{1}{2}\) (x²+y²-2x-15) = 0
= 3(x²+y²)-18x-12y+ 27 = 0

Question 2.
Find the equation and length of the common chord of the circles
S ≡ x² +y²+ 3x + 5y+ 4 = 0 ……………….. (1)
and S ≡ x² +y²+ 5x+ 3y + 4 = 0  …………………. (2)
Solution:
The common chord of two intersecting circles is the radical axis given by S – S’ = 0.
⇒  2x + 2y = 0
⇒ x-y=0 ……………… (3)
Centre of cricle (1) is \(\mathrm{C}_1=\left(-\frac{3}{2},-\frac{5}{2}\right)\) and

∴ AD = 2
∴ Length of the common chord AB = 2(AD) = 2(2) 4

Question 3.
Find the equation of the circle whose diameter is the common chord of the circles
S = x² +y²+2x+3y+ 1=0 …………………. (1)
and S’= x² +y²+4x+3y+2=0 …………………. (2)
Solution:
The common chord is the radical axis of (1) and (2) given by S – S’ = 0.
⇒ 2x-1 =0
⇒  2x+1=0 ………………. (3)
The equation of any circle passing through the point of intersection of (1) and (3) is
S + λL = 0.
(x² +y²+2x+3y+1)+λ(2x+1) = 0
x² +y²+2(λ+ 1)x+3y+(1 +λ) = 0 ……………….. (4)
Centre of this circle = [- (λ + 1), \(-\frac{3}{2}\)]
For the circle (4), 2x + 1 -0 is one chord. This will be the diameter of the circle (4).
If the centre of (4) lies on (3) then
– 2(λ + 1) + 1 = 0
⇒ λ=-\(-\frac{3}{2}\)
∴ The equation of circle whose diameter is the common chord of (1) and (2) is
x² +y² + 2x + 3y + 1) – (2x + 1) = 0
2(x² +y²)+2x+6y+1=0

Question 4.
Find the equation of a circle which cuts each of the following circles orthogonally
S’ = x²+y²+3x+2y+1=0 ……….. (1)
S”=x²+y²-x+6y+5=0 ………… (2)
S”’=x²+y+5x-6y+15=0 …………… (3)
Solution:
Radical axis of circles (1) and (2) is S’ -S”= 0.
⇒ 4x-4y-4=0
⇒ x-y-1=0   ………………… (4)
Radical axis of circles (2) and (3) is S’’ -S”’= 0.
⇒ -6x+ 14y-10 =0
⇒ 3x-7y+5= 0 ……………. (5)
Solving (4) and (5) we get

∴ Radical centre – (3, 2)
Also the length of the tangent from (3, 2) to
S’=\(\sqrt{9+4+9+4+1}=\sqrt{27}\)
Hence equation of circle which cuts orthogonally each of the given circles is obtained by taking radical centre as centre and length of the tangent as radius.
∴ Equation of the required circle is
(x-3)² +(y-2)²= 27
⇒ x²+y²-6x-4y-14=0.

Long Answer Type Questions

Question 1.
Find the equation of the circle which passes through (1, 1) and cuts orthogonally each of the circles
x²+y²-8x-2y+ 16=0 …………. (1)
and x²+y²-4x-4y-1=0 …………. (2)
Solution:
Let the equation of the required circle be
x²+y²+2gx+2fy+c = 0 ………….. (3)
1f this passes through (1, 1) then
1+1 + 2g. 2f + c = 0
⇒ 2g+2f+c- 2 ………….. (4)
If (3) is orthogonal to (1) then
2g(-4) + 2f(-1) = c + 16
=-8g-21=c+16 …………………. (5)
11(3) is orthogonal to (2) then
2g(-2) + 2f(-2) =c-1
= – 4g – 41=c – 1 ………….. (6)
From (5) and (6) we have
– 4g + 21=17
⇒ 4g+2f=-17 …………….. (7)
From (4) and (5) we have
– 6g=+ 14=g= – \(\frac{7}{3}\)
∴ From (7)

Question 2.
Find the equation of circle which is orthogonal to each of the following three circles x²+y²+ 2x + 17y+ 4=0
x²+y²+7x+6y+ 11=0 and x²+y²-x+22y+3=0
Solution:
Denote the given circles by
S=x²+y²+2x+ 17y+4 = 0 ……………….. (1)
S’=x²+y²+7x+6y+11 = 0 ……………… (2)
and S”=x²+y² x+22y+3 = 0 …………………. (3)
The radical axis of S = 0, S’ = 0 is s – S = 0.
⇒ – 5x+11y – 7=0
⇒ 5x – 11y-7=0 ……………….. (4)
The radical axis of S’ = 0. S” = 0 is S’ = S’’ = 0.
⇒ 8x-16y+8=0
⇒ x-2y+1=0 …………… (5)
Solving equations (4) and (5), we get the coordinates of radical centre.

⇒ x = 3, y = 2 ∴ C (3, 2) is the radical centre.
Length of the tangent from C(3, 2) to the circles = 0 is \(\sqrt{9+4+6+34+4}=\sqrt{57}\)
The equation of the circle which cuts or orthogonally each of the three circles if obtained by considering the equation of circle with radical centre (3,2) as centre and length of the tangent \(\sqrt{57}\) as the radius.
∴ Equation of the required circle is
(x-3)² + (y – 2)² = 57
= x² + y²-6x-4y-44 = 0

Question 3.
If the straight line represented by x cos α + y sin α = p intersects the code x² + y² = a² at the points A and B then show that the equation of circle with \(\overline{\mathbf{A B}}\) as diameter is (x²+y²-a²)-2p(x cosα+ysinα – p)=0.
Solution:
Given x cos α + y sin α p …….. (1) intersects the circle
x² + y² = a² ……….. (2) at points A and B.
So the equation of circle passing through the point of intersection of (1) and (2) is
x² + y² = a²+λ(xcos α + ysinα-p) = 0 …………. (3)
Where λ is a parameter.
∴ Centre of (3) is  \(\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)\)
If the circle given by (3) has \(\overline{\mathrm{AB}}\) as diameter then the centre of it must lie on (1). Then

Hence the equation of the required circle is (x² + y²– a²) – 2p(xcosα+ysinα – p)=0.

Question 4.
Show that the circles
S=x²+y²-2x-4y-20 =0 …………….. (1)
and S’=x²+y²+6x – 2y- 90=0 …………… (2)
touch each other Internally. Find their point of contact and the equation of common tangent
Solution:
Let C₁ C₂ be the centres of circles (1) and (2) and r₁ r₂ be the radii of circles (1) and (2).
Then C₁ = (1, 2) and r₁ = \(\sqrt{1+4+20}\) = 5

Since C₁ C₂ – |r₁ r₂|, the two circles touch
Internally. The point of contact P divides C₃ C₄ externally In the ratio of the radii of circles 5: 10 1: 2.

Since the two circles touch internally the common tangent at the point of contact is only the radical axis of
S = 0 and S’- 0 given by S – S’ – 0.
= – 8x – 6y + 70 = 0
= 4x + 3y – 35 = 0

Students must practice this TS Intermediate Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(d) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(d)

I. Find the area of the region enclosed by the given curves.

Question 1.
y = cos x, y = 1 – \(\frac{2 x}{\pi}\)
Solution:
Consider the graphs of the functions
f(x) = cos x and g(x) = 1 – \(\frac{2 x}{\pi}\)

The two curves intersect at (\(\frac{\pi}{2}\), 0) and (π, -1)

∴ The area enclosed between the curves y = cos x, y = 1 – \(\frac{2 \mathrm{x}}{\pi}\) is (2 – \(\frac{\pi}{2}\)).

Question 2.
y = cos x, y = sin 2x, x = 0, x = \(\frac{\pi}{2}\).
Solution:
Consider the graphs of the functions f(x) = cos x and g(x) = sin 2x.

Taking f(x) = cos x and g(x) = sin 2x in [0, \(\frac{\pi}{6}\)]
we have f(x) ≥ g(x) and also in \(\left[\frac{\pi}{6}, \frac{\pi}{2}\right]\) have g(x) ≥ f(x).
Hence area enclosed between f(x) = cos x and g(x) = sin 2x is

∴ Area enclosed between the curves y = cos x, y = sin 2x, x = 0, x = \(\frac{\pi}{2}\) is \(\frac{1}{2}\) sq.units.

Question 3.
y = x³ + 3, y = 0, x = – 1, x = 2. (Mar. ’12)
Solution:

Required area enclosed between y = x³ + 3, y = 0, x = – 1, x = 2 is

Question 4.
y = e^x, y = x, x = 0, x = 1.
Solution:

Question 5.
y = sin x, y = cos x, x = 0, x = \(\frac{\pi}{2}\).
Solution:

Taking f(x) = sin x and g(x) = cos x
We have sin x – cos x < 0 for x ∈ \(\left[0, \frac{\pi}{4}\right]\) and sin x – cos x > 0 for x ∈ \(\left[\frac{\pi}{4}, \frac{\pi}{2}\right]\)
Area enclosed between y = sin x, y = cos x, x = 0, x = \(\frac{\pi}{2}\) is

Question 6.
x = 4 – y², x = 0.
Solution:
The curve x = 4 – y² cuts Y – axis at x = 0.
Hence 4 – y² = 0
⇒ y = ± 2
∴ Points of intersection are given by (0, 2) and (0, – 2).
∴ Area subtended by Y-axis, the curve x = 4 – y² is given by \(A=\int_{-2}^2 x d y=\int_{-2}^2\left(4-y^2\right) d y\)

Question 7.
Find the area enclosed with in the curve |x| + |y| = 1.
Solution:
The given equation of the curve is |x| + |y| = 1 which represents ± x ± y = 1 representing four different lines forming a square.
Consider the line x + y = 1
⇒ y = 1 – x
If the line touches the X-axis then y = 0 and one of the points of intersection with X-axis is (1, 0).
Since the curve is symmetric with respect to coordinate axes, area bounded by |x| + |y| = 1 is

II.

Question 1.
x = 2 – 5y – 3y², x = 0.
Solution:
Solving x = 2 – 5y – 3y² and x = 0
We get – 3y² – 5y + 2 = 0
⇒ 3y² + 5y – 2 = 0
⇒ 3y² + 6y – y – 2 = 0
⇒ 3y (y + 2) – 1 (y + 2) = 0
⇒ (y + 2)(3y – 1) = 0
⇒ y = -2 or y = \(\frac{1}{3}\)
Required area subtended by the curve x = 2 – 5y – 3y², Y-axis and y = \(\frac{1}{3}\) and y = – 2 is

Question 2.
x² = 4y, x = 2, y = 0.
Solution:

The given equation x² = 4y represents parabola which is symmetric with respect to Y-axis.
When x = 2 we have
4y = 4
⇒ y = 1
Hence P(2, 1) is the point of intersection.
∴ Area bounded by x² = 4y, and x = 0, x = 2 is
A = \(\int_0^2 y d x=\int_0^2 \frac{x^2}{4} d x\)
= \(\frac{1}{4}\left[\frac{x^3}{3}\right]_0^2=\frac{1}{4}\left(\frac{8}{3}\right)=\frac{2}{3}\) sq. units.

Question 3.
y² = 3x, x = 3.
Solution:
y² = 3x is a parabola which is symmetrical to X-axis;
when x = 3 then y² = 9
⇒ y = ± 3
Hence (3, 3) and (3, – 3) are the points of intersection.

Question 4.
y = x², y = 2x
Solution:

Given y = x² and y = 2x
solving these we get x² – 2x = 0
⇒ x (x – 2) = 0
⇒ x = 0 or x = 2
∴ y = 0 or y = 4
Hence O(0, 0) and P(2, 4) are the points of intersection of (1) and (2).
∴ Required area bounded by y = x² and y = 2x is
= \(\int_0^2\left(2 x-x^2\right) d x=2\left[\frac{x^2}{2}\right]_0^2-\left[\frac{x^3}{3}\right]_0^2\)
= \(2\left[\frac{4}{2}\right]-\left[\frac{8}{3}\right]\)
= 4 – \(\frac{8}{3}\) = \(\frac{4}{3}\) sq. units.

Question 5.
y = sin 2x, y = √3 sin x, x = 0 and x = \(\frac{\pi}{6}\).
Solution:

Given curves are y = sin 2x …….(1)
and y = √3 sin x ……..(2)
When x = \(\frac{\pi}{6}\) then y = \(\frac{\sqrt{3}}{2}\)
∴ \(\left(\frac{\pi}{6}, \frac{\sqrt{3}}{2}\right)\) is a point of intersection (1) and (2).
∴ Area enclosed between the curves

Question 6.
y = x², y = x³.
Solution:
Given curves are y = x² and y = x³
By solving we get x² = x³
⇒ x² (x – 1) = 0
⇒ x = 0 or x = 1
∴ y = 0 or y = 1
Hence (0, 0) and (1, 1) are the points of intersection.
Also f(x) = x² and g(x) = x³ and f(x) > g(x) in [0, 1].
∴ Area bounded by the curves

Question 7.
y = 4x – x², y = 5 – 2x.
Solution:
The given equations of curves are denoted by
y = f(x) = 4x – x² ……(1)
y = g(x) = 5 – 2x ……..(2)
Solving (1) and (2)
4x – x² = 5 – 2x
⇒ x² – 2x – 4x + 5 = 0
⇒ x² – 6x + 5 = 0
⇒ (x – 5) (x – 1) = 0
⇒ x = 1 or 5
When x = 1, y = 3 and x = 5, y = -5
Hence P(1, 3) and Q(5, – 5) are the two points of intersection.
Now f(x) > g(x) in [1, 5].
∴ Area bounded between the curves is

Question 8.
Find the area in sq.units bounded by the X-axis, part of the curve y = 1 + \(\frac{8}{x^2}\) and the ordinates x = 2 and x = 4.
Solution:
In [2, 4] we have the equation of the curve given by y = 1 + \(\frac{8}{x^2}\)
∴ Area bounded by the curve y = 1 + \(\frac{8}{x^2}\)
X-axis and the ordinates x = 2 and x = 4 is

Question 9.
Find the area of the region bounded by the parabolas y2 = 4x and x2 = 4y.
Solution:
Given equations of curves are
y² = 4x ……..(1)
x² = 4y ……..(2)
Solving (1) and (2) the points of intersection can be obtained.
y² = 4x
⇒ y⁴ = 16x²
⇒ y⁴ = 64y
⇒ y = 4
∴ 4x = y²
⇒ 4x = 16
⇒ x = 4
Points of intersection are (0, 0) and (4, 4).

∴ Area bounded between the parabolas

Question 10.
Find the area bounded by the curve y = log x, the X-axis and the straight line x = e.
Solution:
Area bounded by the curve y = \(\log _e x\), X-axis
and the straight line x = e is \(\int_1^e \log _e x d x\)
= \([\mathrm{x} \log \mathrm{x}]_{\mathrm{1}}^{\mathrm{e}}-\int_1^{\mathrm{e}} \mathrm{dx}\) (∵ when x = e, y = loge e = 1)
= (e – 0) – (e – 1)
= 1 sq.units

III.

Question 1.
y = x² + 1, y = 2x – 2, x = -1, x = 2.
Solution:
Equations of given curves are
y = x² + 1 ……..(1)
y = 2x – 2 ………(2)

Take f(x) = x² + 1 and g(x) = 2x – 1
we find f(x) > g(x) ∀ x ∈ [- 1, 2]
Hence area bounded by y = x² + 1 and y = 2x – 2 is

Question 2.
y² = 4x, y² = 4(4 – x).
Solution:
Equations of the curves are
y² = 4x ………(1)
y² = 4(4 – x) ………(2)
From (1) and (2)
4x = 4(4 – x)
⇒ 8x = 16
⇒ x = 2
∴ y² = 8
⇒ y = ± 2√2
∴ The points of intersection are (2, 2√2), (2, – 2√2)

Question 3.
y = 2 – x², y = x².
Solution:
The given curves are
y = 2 – x² ………(1)
y = x² …….(2)

Solving (1) and (2) we get
2 – x² = x²
⇒ 2x² = 2
⇒ x² = 1
⇒ x = ± 1
∴ y = 1
Hence points of intersection of curves are (1, 1), (- 1, 1)
∴ Area bounded between the curves

Question 4.
Show that the area enclosed between the curves y² = 12 (x + 3) and y² = 20 (5 – x) is 64 \(\sqrt{\frac{5}{3}}\).
Solution:

Given curves are y² = 12 (x + 3) …….(1)
and y² = 20 (5 – x) ………(2)
Solving (1) and (2) we get
12 (x + 3) = 20 (5 – x)
⇒ 12x + 36 = 100 – 20x
⇒ 32x = 64
⇒ x = 2
∴ y² = 20(5 – 2) = 60
⇒ y = ± 2√15
∴ Points of intersection are B(2, 2√15) and B'(2, – 2√5).
The given equations represent these parabolas
(y – β)² = 4a(x – α) and
(y – β)² = – 4a(x – α)
∴ Vertex of (1) is A(- 3, 0) and vertex of (2) is A'(5, 0).
∴ The required area is symmetrical about X-axis.
∴ Area bounded by curves is ABA’B’

Question 5.
Find the area of the region {(x, y): x² – x – 1 ≤ y ≤ -1}.
Solution:
Consider y = x² – x – 1 ……(1)
and y = – 1 ……(2)

Which represents a parabola in the form (x – h)² = 4a(y – k) where vertex \(\left(\frac{1}{2}, \frac{-5}{4}\right)\) with 4a = 1.
Also x² – x – 1 = -1
⇒ x² – x = 0
⇒ x (x – 1) = 0
⇒ x = 0 or 1
∴ Area subtended by the region {(x, y) : x² – x – 1 ≤ y ≤ -1}

∴ Area of the region {(x, y); x² – x – 1 ≤ y ≤ – 1} is \(\frac{1}{6}\).

Question 6.
The circle x² + y² = 8 is divided into two parts by the parabola 2y = x². Find the area of both the points.
Solution:

Given curves are x² + y² = 8 ……(1)
and 2y = x² …….(2)
Solving (1) and (2)
x² + \(\frac{x^4}{4}\) = 8
⇒ x⁴ + 4x² – 32 = 0
⇒ x⁴ + 8x² – 4x² – 32 = 0
⇒ x² (x² + 8) – 4(x² + 8) = 0
⇒ (x² – 4) (x² + 8) = 0
⇒ x = ±2 (∵ x² + 8 = 0 is not admissible)
∴ y = 2
Hence the points of intersection of (1) and (2) are (2, 2) (- 2, 2).
Let the area bounded between (1) and (2) be sum of two areas A₁ and A₂ as shown in the figure. Then

Again
A₂ = π + \(\frac{2}{3}\) (Since the bounded position is symmetric)
∴ Area bounded by the curves = 2π + \(\frac{4}{3}\) sq.units
Area bounded by the circle = πr² = 8π when r = 2√2
∴ Area of the remaining portion = 8π – (2π + \(\frac{4}{3}\)) = 6π – \(\frac{4}{3}\) sq.units.

Question 7.
Show that the area of the region bounded by \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 (ellipse) is πab. Also deduce the area of the circle x² + y² = a². (June ’10)
Solution:

Given equation of ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
Since the ellipse is symmetric with respect to coordinate axes we have

∴ Area of the region bounded by \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 is πab sq.units
Substituting b = a in the above result we have the area of circle given by πa(a) = πa² sq.units

Question 8.
Find the area of the region enclosed by the curves y = sin πx, y = x² – x, x = 2.
Solution:
The graphs of the given equations y = sin πx and y = x² – x, x = 2 are shown below.

Question 9.
Let AOB be the positive quadrant of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 with OA = a, OB = b. Then show that the area bounded between the chord AB and the arc AB of the ellipse is \(\frac{(\pi-2) a b}{4}\).
Solution:

Question 10.
Prove that the curves y² = 4x and x² = 4y divide the area of square bounded by the lines x = 0, x = 4, y = 4 and y = 0 into three equal parts.
Solution:

The given curves are y² = 4x ……(1) and x² = 4y ………(2)
Solving y⁴ = 16x² = 64y
⇒ y(y³ – 64) = o
⇒ y = 0 or y = 4
When y = 4 we have 4x = 16 ⇒ x = 4
∴ Points of intersection of parabolas is P(4, 4).
∴ Area bounded by the parabolas

Area of the square formed = (OA)² = (4)² = 16
Since the area bounded by the parabolas x² = 4y and y² = 4x is \(\frac{1}{2}\) sq.units.
Which is one third of the area of square we conclude that the curves y² = 4x and x² = 4y divide the area of the square bounded by x = 0, x = 4, y = 0, y = 4 into three equal parts.

Students must practice these TS Inter 2nd Year Maths 2B Important Questions Chapter 1 Circles to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2B Circles Important Questions

Very Short Answer Type Questions      

Question 1.
Find the equation of circle with centre (1, 4) and radius 5.
Solution:
Standand equation of circle with centre
(h. k) and radius ‘r’ is (x – h)² + (y – k)² r²
(h, k) = (1, 4) and r= 5
∴ Equation of circle is (x – 1)² + (4)² = 25
x². y² -2x – 8y+ 17=25
= x²+y²– 2x – 8y – 8 = 0

Question 2.
Find the centre and radius of the circle
x² + y² + 2x – 4y – 4 = 0.
Solution:
Comparing with x²+y²+2gx+2fy+c = 0
2g=2, 21= – 4, and c = – 4
∴ Centre=(-g,-f)=(-1,2) and
∴ Radius = \(\sqrt{g^2+f^2-c}=\sqrt{1+4+4}=3\)

Question 3.
Find the centre and radius of the circle
3x² + 3y² – 6x + 4y – 4 = 0.
Solution:
Given equation can be written as

Question 4.
Find the equation of the circle whose centre is (-.1, 2) and which passes through (5,6).
Solution:
Let C (-1, 2) be the centre of the circle. Since (5, 6) is a point on the circle, the radius of the circle.

Question 5.
Find the equation of circle passing through (2,3) and concentric with the circle
x² + y² + 8x + 12y + 15 = 0.
Solution:
Let the equation of required circle be
x² + y² + 2gx + 2fy + c‘ = 0
If It passes through (2, 3) then
4+9+8(2)+ 12(3) + c‘ = 0
= 65 + c’ = 0 ⇔ c’ = – 65
∴ The equation of required circle is x² + y² + 8x + 12y – 65 = 0.

Question 6.
If the circle x² + y² +ax+by -12 = 0 has the centre at (2,3) then find a, b and the radius of the circle.
Solution:
The equation of circle is
x² + y² +ax+by -12 = 0
Centre 01 the circle = \(\left(\frac{-\mathrm{a}}{2},-\frac{\mathrm{b}}{2}\right)\) = (2, 3) (given)
∴ a =- 4 and b = – 6
∴ Radius of the circle = \(\sqrt{4+9+12}=5\)

Question 7.
If the circle x² + y² – 4x+6y+a = 0 has radius 4 then lead ‘a’.
Solution:
Given equation of circle is
x² + y² – 4x + 6y+a = 0,
centre C = (2, -3) and radius = 4 (given)
∴ \(\sqrt{4+9-\mathrm{a}}\) = 4 ⇒ 13 – a = 16 ⇒ a – 3.

Question 8.
Find the equation of the circle whose extremities of diameter are (1, 2) and (4, 5).
Solution:
Taking A(1, 2) (x₁,y₁) and B(4, 5) (x₂,y₂) the equation of circle having A, B as extremities of diameter is
(x-x₁)(x-x₂)+(y-y₁)(y-y₂)=0
(x – 1)(x-4) + (y-2)(y-5) =0
= x²-5x+4+y²-7y+ 10 =0
= x²+y²-5x-7y+ 14=0

Question 9.
Find the other end of the diameter of the circle x² + y²-8x – 8y + 27 = 0 if one end of it is  (2, 3).
Solution:
Centre of the given circle is C (4, 4).
One end of diameter is A = (2, 3). Let the other end be B (x, y). Then C is the end point of AB.
∴ \(\frac{x+2}{2}=4 \) and \( \frac{y+3}{2}=4\)
⇒ x = 6, y = 5
∴ Other end of the diameter B = (6, 5)

Question 10.
Obtain the parametric equations of x² + y² = 1.
Solution:
Centre of the circle = (0, 0) and radius = 1 = (h, k)
The parametric equations of curve are
x = h + rcosθ = 0 + 1. cosθ = cosθ
y = k + rsinθ = θ+1.sinθ = sinθ
0 ≤ θ ≤ 2π

Question 11.
Obtain the parametric equation of the circle represented by x² + y² + 6x + 8y – 96 = 0.
Solution:
Centre(h,k) =(-3,-4)
and radius r = \(\sqrt{9+16+96}=\sqrt{121}\) = 1
∴ x = h + r cosθ = – 3 + 11 cosθ
y = k + r sin 0=- 4 + 11 sin θ,0≤0≤ 2π

Question 12.
Locate the position of the point (2, 4) w.r.t circle x² + y² – 4x – 6y + 11 = 0.
Solution:
Here (x, y) (2, 4) and
x² + y² – 4x – 6y+ 11=0
S ≡ (2)²+(4)²– 8 – 12 +11=-1
Since S₁₁ <0, the point (2, 4) lies Inside the circle.

Question 13.
Find the length of the tangent from (1,3) to the circle x² + y² -2x + 4y – 11 = 0.
Solution:
Given (x₁, y₁) = (1, 3) and
S ≡ x²+y²-2x+4y-11 = 0
S₁₁ = 1²+3² -2 + 12 – 11 = 9
∴ Length of the tangent from P(x₁, y₁) to S = 0 is
= \(\sqrt{S_{11}}=\sqrt{9}\) = 3

Question 14.
Show that the circle S ≡ x² + y²+2gx+2fy+ c = 0 touches (i) x-axis if g² = c (ii) Y-axis if f² =c.
Solution:
(i) We have the Intercept made by S = 0 on X-axis is \(2 \sqrt{\mathrm{g}^2-\mathrm{c}}\).
If the circle touches X-axis then \(2 \sqrt{\mathrm{g}^2-\mathrm{c}}\) = ⇒ g²= c.

(ii) Similarly if the intercept made by S = 0 on
Y-axis is \(2 \sqrt{f^2-c}\) . If the circle touches Y-axis then \(2 \sqrt{f^2-c}\) = f²=c.

Question 15.
Find the equation of tangent to x² + y² – 6x +4y – 12 = 0 at (-1,1).
Solution:
We have the equation of tangent at (x₁, y₁) to
S = 0 is xx₁ +yy₁ +g(x+x₁)+1(y+y₁)+ c
⇒ x(-1) 4y(1) – 3(x-1) + 2(y+ 1)-12 = 0
⇒ – 4x+3y-7 = 0 ,
⇒ 4x – 3y+ 7 = 0

Question 16.
Show that the line 5x + 12y – 4 = 0 touches the circle x² + y² – 6x + 4y + 12 = 0.
Solution:
Centre of the given circle = (3, -2) and
radius = \(\sqrt{9+4-12}=1\)
The perpendicular distance from the centre
(3,-2) to the line 5x + 12y- 4 = 0 is
\(=\left|\frac{5(3)+12(-2)-4}{\sqrt{25+144}}\right|=\left|\frac{-13}{13}\right|=1\)
∴ radius of the circle.
⇒ The line 5x + 12y-.4 = 0 touches the given circle.

Question 17.
Find the area of the triangle formed by the tangent at P(x₁, y₁)to the circle x² + y² = a² with the coordinate axes where x₁ y₁ ≠ 0.
Solution:
Equation of tangent at (x₁, y₁) to the circle
x² +y²-a² is xx₁ +yy₁– a²=0.
x, y intercepts are \(\frac{a^2}{x_1}\) and \(\frac{\mathrm{a}^2}{\mathrm{y}_1}\)
∴ Required area of the triangle
=\(\frac{1}{2}\left|\frac{a^2}{x_1} \cdot \frac{a^2}{y_1}\right|=\frac{a^4}{2\left|x_1 y_1\right|}\)

Question 18.
State the necessary and sufficient condition forlx+ my + 0 to be normal to the circle x²+ y² + 2gx + 2fy + c = 0.
Solution:
The straight line lx + my + n = 0 is a normal to the circle S ≡ x² + y² + 2 + 2fy + c = 0.
⇔ Centre (-g, – f) of the circle lies on lx + my + n = 0
⇔ l(-g) + m(-f) + n = 0
⇔ lg + mf = n

Question 19.
Find the condition that the tangents are drawn from the exterior point (g,f) to S ≡ x²+y²+ 2gx + 2fy + c= 0 are perpendicular to each other.
Solution:
If the angle between the tangents drawn from P(x₁,y₁) to S=0 is θ then

Question 20.
Find the chord of contact of (2, 5) with respect to the circle x² + y² – 5x + 4y -2 = 0.
Solution:
2g=-5 and 2f = 4 ⇒ g\(\frac{5}{2}\) and f=2,c=-2
Equation of chord of contact of (x₁, y₁) w.r.t S = 0 is
xx₁ +yy₁ + g(x+x,) +f(y+y₁)+c=0
=2x+5y-(x+2)+2(y+5)-2=0
= x-14y+6=0

Question 21.
Find the equation of the polar of the point (2, a)w.r.tthe circle x²+y²+6x +8y-96 =0.
Solution:
Equation of polar of (x₁, y₁) (2, 3) is +yy +g(x+x₁)+f(y+y₁)+c=0
⇒ x(2) +y(3)+ 3(x + 2) +.4(y+ 3)- 96 = 0
⇒ 5x + 7y – 78 = 0
⇒ (x₁– a)²= (x₁ +a)²+y²
⇒ (x₁ – a)² – (x₁ + a)² y²₁
⇒ y – 4ax₁ ⇒ y²₁ + 4ax₁ = 0
∴ Locus of (x₁, y₁) is y² + 4ax = 0.

Question 22.
Find the pole of the line x+y+2 = 0 w.r.t x² + y²– 4x + 6y – 12=0.
Solution:
Here lx +my+n=0 is x+y+2=0 and S=0 is x² + y² – 4x + 6y – 12 = 0.

Question 23.
Show that (4, -2) and (3, -6) are conjugate w.r.t. the circle x² + y² – 24 = 0.
Solution:
Here (x₁, y₁) = (4, – 2) and (x₂, y₂) (3, -6) and S = x² + y²-24 = 0 ……….. (1)
Two points (x₁, y₁) and (x₂, y₂) are conjugate w.r.t S=0 if S₁₂=0
∴ x₁x₂+y₁y₁-24 = 0
For the given points
S₁₂ =4(3)+(-2)(-6)-24-0
∴ The given points are conjugate w.r.t the given circle.

Question 24.
If (4, k) and (2,3) are conjugate points w.r.t x² + y² = 17 then find k.
Solution:
(x₁, y₁)= (4, k) and (x₂, y₂)= (2, 3). Since the given points are conjugate S₁₂ = 0.
= x₁x₂ + y₁y₂ – 17 = 0
(4)(2)+(k)(3)-17=0 ⇒ k=3

Question 25.
Show that the lines 2x+3y+ 11 =0,and 2x – 2y -1= 0 are conjugate w.r.t x² + y²+ 4x + 6y + 12 = 0.
Solution:

Question 26.
Find the inverse point of (2, -3) wrt the circle x²+y²– 4x-6y+9=0
Solution:
Let P(-2, 3) and C (2,3) is the centre of the given circle. Then the polar of P is
x(-2)+y(3)-2(x-2)-3(y+3)+9=0
x = 1 ……………. (1)
Equation of line \(\overline{\mathrm{CP}}=\mathrm{y}-3=\frac{3-3}{2+1}(\mathrm{x}+2)\)
⇒ y – 3 = 0 ⇒ y = 3 ……………. (2)
∴ From (1) and (2) the inverse point of
P(-2, 3) is (1, 3).

Short Answer Type Questions

Question 1.
From the point A(0, 3) on the circle x² + 4x +(y-3)²= 0 a chord AB is drawn and extended to a point M such that
AM = 2AB. Find the equation to the locus of M.
Solution:

Let M (x₁, y₁) be the locus. Given AM – 2A8
= AB+ BM = AB+AB
BM – AB ⇒ B is the mid point of AM

Question 2.
Find the equation of the circle passing through (4, 1), (6,5) and having the centre on the line 4x+y-16=0.
Solution:
Let the equation of the required circle be
x²+y²+2gx+2fy+c = 0 ……………. (1)
Since it passes through (4, 1) we have
16 + 1 + 8g + 2f + c = 0
= 17+8g+2f+c=0 ……………. (2)
Similarly (6, 5) lies on (1) then
36+25+12g+ 10f +c=0
= 61+12g+10f+c=0 ……………. (3)
Given that the centre of circle (-g, -f) lies on 4x + y-16 = 0
-4g-f-16=0
⇒ 4g+f+16=0 ……………. (4)
From (2) and (3)
– 44 – 4g – 8f 0
=g+2f=- 11  ………….. (5)
From (4) 4g+ f =- 16

⇒ g= – 3, f =- 4 and
from (2)
17 – 24 – 8+c – 0 ⇒ c = 15
∴ Equation of the required circle from (1) is
x² +y² – 6x – 8y + 15=0.

Question 3.
Suppose a point (x₁, y₁) satisfies x² + y² + 2gx + 2fy + c= 0 then show that it represents a circle whenever g, f and c are real.
Solution:
Comparing the given equation with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we have coefficient of x² = coefficient of y² and coefficient of xy term = 0.
The given equation represents a circle If g² + f² ≥ 0
Since (x₁, y₁) is a point on the circle we have
x₁ + y₁ + 2gx₁ + 2fy₁ + c = 0
g² + f² – c = g² + f² +x²₁ +y²₁ + 2gx₁ + 2fy₁ = (x₁ +g) + (y₁ + f)² ≥ 0
Since g, f and c are real the equation (1) represents a circle.

Question 4.
Find the equation of circle which Louches x-axis at a distance of 3 from the origin and making intercept of length 6 on y-axis.
Solution:

Let the equation of required circle be
x²+y²+2gx+2fy+c=0 …………………….. (1)
If it touches x- axis at (3, 0) then 9 + 0 + 6g + c = 0
⇒ 6g+c= – 9 …………………… (2)
If circle touches x-axis then g² – c = 0 ………………………. (3)
Adding (2) and (3)
g² + 6g = -9
= (g+3)² = 0 = g = -3 ……………………. (4)
∴ From (3), C = 9
Also given that intercept on y-axis is 6

Question 5.
Find the equation of circle which passes through the vertices of the triangle formed by
L₁ =x+y+ 1 =0, L₂=3x+y-5=0 and L₃ = 2x + y-5 = 0.
Solution:
Suppose L₁, L₂; L₂, L.₃ and L₃, L₁ intersect at A, B and C respectively. Consider a curve whose equation is
k(x+y+1)(3x+y-5)+1(3x+y-5)
(2x+y-5)+m(x+y+ 1)
(2x+y-5) = 0 ……………………. (1)
We can verify that this curve passes through A, B, C. So we find k, I and m such that the equation (1) represents a circle. If (1) represents a circle then
(i) coefficient of x² = coefficient of y²
= 3k + 6l + 2m = k + l + m
= 2k+5l+m=0 ………………… (2)
(ii) coefficient of xy is zero.
4k+5l+3m= 0 ………………… (3)
Solving (2) and (3) we get

Hence the required equation is
5(x+y+1)(3x+y-5)-1(3x+y-5)
(2x+y-5)-5(x+y+1)
(2x + y -5) = 0
⇒ x²+y²-30x-10y+ 25=0

Question 6.
Find the centre of the drive passing through the points (0,0), (2,0) and (0,2).
Solution:
Let the equation of required circle be
x²+y²+2gx+2fy+c=0 ……………….. (1)
If (1) passes through (0, 0) then c = 0
If (1) passes through (2, 0) then
4+4g+c=0 ………….. (1)
If (1) passes through (0, 2) then
4+4f+c=0 ………….. (2)
From (2) and (3) we have
g=-1 and f = – 1 (∵ c=0)
∴ Centre of the circle = (-g, -f) (1, 1)

Question 7.
If a point P is moving such that the length of tangents drawn from P to x²+y² – 2x + 4y – 20 = 0
x²+y² – 2x-8y+ 1=0 are in the ratio 2: 1 then show that the equation of the locus of P is x²+y² -2x – 12y+8=0.
Solution:
Let P (x₁, y₁) be the locus and \(\overline{\mathrm{PT}_1}, \overline{\mathrm{PT}_2}\) are the tangents drawn from the points P to the two circles x²+y² – 2x + 4y – 20 = 0 and x²+y²– 2x – 8y + 1 = 0

Question 8.
lf S≡ x²+y²+2gx+2fy+c=0 represents a circle then show that the straight line lx + my + n = 0.
(i) touches the circle S = 0 if
\(\left(g^2+f^2-c\right)=\frac{(g l+m f-n)^2}{\left.l^2+m^2\right)}\)

(ii) meets the circle S=0 in two points if.
\(g^2+f^2-c>\frac{(g l+m f-n)^2}{\left.a^2+m^2\right)}\)

(ii) will not meet the circle if
\(\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}<\frac{(\mathrm{g} l+\mathrm{mf}-\mathrm{n})^2}{\left.l^2+\mathrm{m}^2\right)}\)
Solution:
(i) The given straight line lx + my + n = 0
touches the circle S ≡ x²+y² + 2gx + 2fy + c = 0 if the perpendicular distance from (-g, -f) to lx + my + n – 0 is equal to radius r.

(ii) The given line meets the circle S=0 in two points
\(\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}>\frac{(\mathrm{lg}+\mathrm{mf}-\mathrm{n})^2}{l^2+\mathrm{m}^2}\)

(iii) The given line will not meet the circle S=0
If \(\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}<\frac{(\mathrm{gl} l \mathrm{mf}-\mathrm{n})^2}{l^2+\mathrm{m}^2}\)

Question 9.
Find the length of the chord intercepted by the circle x²+y²+8x-4y – 16 = 0 on the line 3x-y+4 = 0.
Solution:

Centre of the circle C = (-4, 2) and
radius= \(\sqrt{16+4+16}=6\)
CL = Perpendicular distance from C(-4,2) to the chord 3x-y + 4=0.

Question 10.
Find the equation of tangents to x²+y²-4x+6y-12=0 which are parallel to x + 2y -8 = 0.
Solution:
Centre of the given circle C = (2, -3) and radius \(\sqrt{4+9+12}\) =5
Any line parallel to x + 2y – 8= 0 is of the form x + 2y + k – 0. If this line becomes a tangent then the perpendicular distance from C(2, -3) to x + 2y + k = 0 is equal to the radius.
∴ \(\left|\frac{2-6+k}{\sqrt{1+4}}\right|=5\)
⇒ |k – 4| = 5\(\sqrt{5}\) ⇒ k = 4 ± 5\(\sqrt{5}\)
∴ Equation of parallel tangents are
x+2y+(4±5\(\sqrt{5}\)  )=0

Question 11.
Find the equation of tangent to x²+y² – 2x+ 4y = 0 at (3, -1). Also find the equation of tangent parallel to it.
Solution:
Equation of tangent at (3, -1) to the circle
x²+y² -3x + = 0 is
x(3)+y(-1)-1(x+3)+2(y-1)=0
= 3x-y-1(x+ 3) + 2(y-1) 0
3x-y-x-3+2y-20
2x+y-5=0 ……………………. (1)
Equational line parallel to 2x + y-5 = 0 is of the from 2x + y + k = 0. If this is a tangent to the given circle then the perpendicular distance from the centre (1, -2) is equal to the radius=\(\sqrt{1+4}=\sqrt{5}\)
∴ \(\left|\frac{2(1)-2+k}{\sqrt{5}}\right|=\sqrt{5}\)
= k -±5
∴ Equations of parallel tangents to (1) are 2x + y ± 5 = 0
∴ The equation of other parallel tangent is 2x + y + 5 = 0

Question 12.
If 4x-3y+7=0 is a tangent to the circle represented by x²+y²-6x+4y-12=0 then find the point of contact.
Solution:
Centre of the given circle C = (3, -2). Let the P(x₁, y₁) be the contact.
Then 4x₁ – 3y₁ + 7 = 0 …………… (1) is perpendicular to PC, equation of PC is 3x + 4y + k = 0
Since this passes through C(3, -2) we have
9 – 8+k=0=k=-1
∴ Equation of CP is 3x₁ + 4y₁ – 1 = 0 ……………. (2)
Solving (1) and (2)

∴ P(-1, 1) is the point of contact.

Question 13.
Find the equations of circles which touch 2x – 3y+ 1 =0 at (1, 1) and having radius \(\sqrt{13}\)
Solution:

Equation of line ⊥r to 2x – 3y 1 – 0 is of the from 3x + + k -0; Since this passes through
(1, 1) we have 3+2 +k=0 ⇒ k=-5
Equation of line perpendicular to the tangent is 3x+2y-5=0 ………….. (1)
Let (x, y) be the centre of circle

⇒ x² – 2x-3=0 ⇒ (x-3)(x+1)=0
⇒ x = 3 or x = – 1
When x = 3, we have from (1) y =\(\frac{5-9}{2}\) – 2
and when x=-1, y = \(\frac{5+3}{2}=4\)
∴ Centre are (3, -2) and (-1, 4).
∴ Equations of circles with (3, – 2) and
(-1, 4) With radius ,\(\sqrt{13}\) are given by
(x-3)² + (y+ 2)² = 13 and (x+ 1)² + (3,4)² = 13
= x² +y²-6x+4y=0 and x² +y²-2x-8y+4 = 0.

Question 14.
Find the equation of the tangent at the point 300 (Parametric value of O) of the circle x2 4-y2+4x+6y-39=0.
Solution:
Here g=2, f=3. c=-39
\(r=\sqrt{4+9+39}=\sqrt{52}=2 \sqrt{13}\)
The required equation of tangent at ‘θ’ to
S = 0 is given by the formula

Question 15.
Find the equation of normal to the circle x²+y²– 4x-6y+ 11 =0 at (3, 2), Also find the other point where the normal meets the circle.
Solution:
Let A(3, 2) and C be the centre of the given circle C = (2, 3) = (-g,-f)
Equation of normal at (x₁, y₁) is (x-x₁)(y₁+f)-(y-y₁)(x₁+g) = 0
⇒ (x-3)(2-3)-(y-2)(3-2)=0
⇒ 1(x-3)- 1(y-2)= 0
⇒ x-y+5 =0 x+y-5 =0
Let B (x₁, y₁) be the other point where the normal meets the circle.
Then \(\frac{x_1+3}{2}=2\) and \(\frac{y_1+2}{2}=3\)
x₁ = 1, and y₁ = 4
Hence normal at (3,2) meets the circle at (1,4).

Question 16.
Find the area of the triangle formed by the normal at (3, -4) to the circle x²+y²-22x-4y+ 25=0 with the coordinate axes.
Solution:
From the given equation of circle
2g = -22 and 2f = – 4=g=-11 and f=-2
Also (x₁, y₁) =(3,-4)
Then equation of normal al (x₁, y₁) is
(x-x₁)(y-y₁)-(y-y₁)(x₁ +g)=0
⇒ (x-3)(-4-2)-(y+4)(3-11)=0
⇒ (x-3)(-6)-(y+4)(-8)=0
⇒ -6x + + 50 = 0
⇒ 3x – 4y – 25 = 0

Question 17.
If θ₁, θ₂ are the angles of inclination of tangents through a point P to (lie circle x² + y²= a² then find the locus of P where cot θ₁+ cot θ₂ = k
Solution:
The equation of tangent to the circle x² + y² = a² having slope m is y = mx + \(a \sqrt{1+m^2}\)
Let P(x₁, y₁) be a point on the locus. Then

Question 18.
If the chord of contact of P with respect to the circle x² + y² = a² cut the circle at A and B such that \(\angle \mathrm{AOB}=90^{\circ}\) then show that P lies on the circle x²+y²=2a².
Solution:

Let P(x₁, y₁) be a point and let the chord of contact of P(x₁, y₁) meets circle are A and B.Such that \(\angle \mathrm{AOB}=90^{\circ}\)
Equation of chord of contact of P(x₁, y₁) is
xx₁ + yy₁ – a₂ = 0 ⇒ \(\frac{x_1+y y_1}{a^2}=1\) ……….. (1)

Question 19.
Show that the poles of tangents to the circle x+y = a² w.r.t. the circle (x + a)² + 2 = 11 on y² + 4ax = 0.
Solution:
Let P(x₁, y₁) be the pole of the tangent to the circle x² + y² – a² …………………… (1) w.r.t circle (x + a)² + y² = 2a². Then the equation of polar of P (x₁, y₁) w.r.t
(x + a)² + y² . 2a² is …………………. (2)
xx₁ +yy₁ +a(x₁+ y₁)-a²= 0
= x(x₁ + a) + yy₁ + (ax₁ – a₂) = 0 ………………. (3)
The line is a tangent to the circle (1) then perpendicular distance from (0, 0) to (3) is equal to radius ‘a’.

Question 20.
Show that the area of the triangle formed by the two tangents through P(x₁, y₁) to the circle S=x²+y²+2gx+2fy+c=0 and the chord of contact of P.w.r. IS=0 is \(\frac{r\left(S_{11}\right)^{\frac{3}{2}}}{S_{11}+r^2}\)  where r is the radius of the circle.
Solution:

Let PA and PB be the two tangents drawn from P(x), y) to the circle S = 0 and θ be the angle between these two tangents.
Then tan = \(\frac{\theta}{2}=\frac{r}{\sqrt{S_{11}}}\)
Area of triangle formed by the tangents through P(x₁, y₁) to S = 0 and the chord of contact of P w.r.t S = 0

Question 21.
Find the mid point of the chord Intercepted by x²+ y² – 2x – 10y + 1 = 0 on the line x – 2y + 7 = 0.
Solution:
Let x²+ y² – 2x – 10y + 1 = 0 ………………. (1)
x – 2y+ 7= 0 ……………… (2)
Let P(x₁, y₁) be the midpoint of the chord intercepted by the circle (L) on the line given by (2).
The equation of chord of (x₁, y₁) in terms of its midpoint is

Question 22.
Find the equation of pair of tangents drawn from (10, 4) to the circle x² + y² = 25.
Solution:
Equation of pair of tangents is S. S₁₁ = S₁²
⇒ (100+16-25)(x²+y²-25)=(10x+4y-25)²
⇒ 91(x²+ y²-25)=100x²+16y²+625+80xy – 200y – 500x
⇒ 9x² +80xy-75y²-500x-200y + 2900=0

Long Answer Type Questions

Question 1.
Find the equation of circle passing through P(1, 1), Q(2, -1) and R(3, 2).
Solution:

Question 2.
Find the equation of the circumcircle of the triangle formed by the line ax + by + c=0 (abc ≠ 0) and the coordinate axes.
Solution:
Let the line ax+by+c=0 cuts x, and y axis at A and B so that

are the vertices of the triangle.
Let x²+y²+2gx+2fy+c=0 …………… (1) be the required equation of the circle. Since it passes through (0, 0) we have c= 0.

Question 3.
Find the locus of mid points of the chords of contant of x² + y² = a² from the points lying on the line lx + my + n=0.
Solution:
Let (x₁, y₁) be the locus of mid points of chords of the circle x² + y² = a² ………………. (1)
and this is a chord lies on ix + my + n = 0 ………………… (2)
i.e., pole of this chord is on (2).
Equation of chord of (1) having (x₁, y₁) as its mid point is xx₁ + yy₁ – x + y

Question 4.
Show that four common tangents can be drawn for the circles given by
x²+y²-14x+6y+33=9 ……………… (1)
x²+y²+30x-2y+1=0………………(2)
and find the Internal and external centres of similitude.
Solution:
Centre of circLe (1) is C₁ = (7, – 3)
Centre of circle (2) is C₂ = (15, 1)

Question 5.
Prove that the circles
x²+y²-8x-6y+21=0 ……………. (1) and x²+y²-2y-15=0 ………………… (2) have exactly two common tangents.
Also find the point of Intersection of those tangents.
Solution:
Centres of circles are C₁ (4, 3) and C₂ – (0, -1)

∴ Given circles intersect each other and have exactly two common tangents.
r₁ : r₂ = 2 : 4 = 1: 2
The point of contact P divides C₁C₂ externally in the ratio 1: 2.
∴ External centre of similitude
= \(\left(\frac{8-0}{2-1}, \frac{6-1}{2-1}\right)=(8,5)\)

Question 6.
Show that the circles
x²+y²– 4x-6y-12=0 ………………. (1)
and x²+y²+6x+18y+26=0 …………….. (2) touch each other. Also find the point of contact and common Langent at this point of contact.
Solution:
Let C₁ (2, 3) and C₂ (- 3, -9) are centres of circles (1) and (2) and their radii are

C₁C₂ = r₁ + r₂ and hence the two circles touch each other externally. Point of contact divides C₁C₂ in the ratio
r₁ : r₂ = 5 : 8

Question 7.
Show that the circles x²+y²-4x-6y-12=0 and 5(x²+y²)-8x-14y-32=0 touch each other and find their point of contact.
Solution:

Now \(\overline{\mathrm{C}_1 \mathrm{C}_2}\) = | r₁-r₂ |
Here the circles (1) and (2) touch each other internally the point of contact P dividies C₁C₂ in the ratio 5 : 3 externally.
\(\mathrm{P}=\left(\frac{3(2)-5\left(\frac{4}{5}\right)}{3-5}, \frac{3(3)-5\left(\frac{7}{5}\right)}{3-5}\right)\)
E (-1, – 1)
∴ Point of contact = (-1, -1)

Question 8.
Find the equations to all possible common tangents of the circles
x²+y² -2x-6y+6=0 ………………. (1) and x²+y² = 1 …………………. (2)
Solution:
Centres of circles are C₁ – (1, 3) and C₂ (0, 0)

= 4y + 3xy- 9y – 3x + 5 = 0
= (y +l) (4y + 3x + m) (Suppose)
Equating the coefficient of x, y and constant terms
3l = – 3 ………………. (3)
and 4l + m = – 9 ……………..(4)
lm=5 …………….. (5)
From (3) and (4) l=-1⇒m=-5
Equations of transverse common tangents are (y-1) = 0 and 4y+3x-5=0
Direct common tangents are given by
(x²+y²-1)(1+9-1)=(xi-3y+ 1)²
=9(x²+y²-1)=x²+9y²+1+6xy+6y+2x
8x²– 6xy-2x – 6y- 10=0
= (x+l) (8x-6y+m)
Comparing coefficient of x, y and constant and
8l+m=-2
and -6l=-6 =l= l and
lm = – 10 ⇒ m – 10
Equations of direct common tangents are
x+ 1-0 and 8x-6y-10-0
⇒ 4x – 3y – 5 = 0