Let the eccentricity of the hyperbola <math> <mrow> <mi>H</mi> <mo>:</mo> <mfrac> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <msup> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> <mo>−</mo> <mfrac> <mrow> <msup> <mrow> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mtext> </mtext> <mtext> </mtext> <mi>b</mi> <mi>e</mi> <mtext> </mtext> <mtext> </mtext> <mroot> <mrow> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mrow></mrow> </mroot> </mrow> </math> and length of its latus rectum be <math> <mrow> <mn>6</mn> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mi>i</mi> <mi>f</mi> <mtext> </mtext> <mtext> </mtext> <mi>y</mi> <mo>=</mo> <mn>2</mn> <mi>x</mi> <mo>+</mo> <mi>c</mi> </mrow> </math> is a tangent to the hyperbola H, then the value of c2 is equal to

a 2 ( e 2 − 1 ) = b 2 e = 5 2 ⇒ b 2 = 3 a 2 2 Length of latus rectum 2 b 2 a = 6 2 3 a = 6 2 ⇒ a = 2 2 b = 2 3 y = 2x + c is tangent to hyperbola ∴ c 2 = a 2 m 2 − b 2 = 2 0

Ask & Answer Question

Maths NCERT Exemplar Solutions Class 11th Chapter Seven Conic Sections Ncert Solutions Maths class 11th Maths Class 11th

Let the eccentricity of the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 b e \sqrt[]{\frac{5}{2}}$ and length of its latus rectum be $6 \sqrt[]{2}, i f y = 2 x + c$ is a tangent to the hyperbola H, then the value of c² is equal to

Option 1 -

Option 2 -

Option 3 -

Option 4 -

2 Views | Posted 2 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

2 weeks ago

Correct Option - 2

Detailed Solution:

$a^{2} (e^{2} - 1) = b^{2}$

$e = \sqrt[]{\frac{5}{2}} \Rightarrow b^{2} = \frac{3 a^{2}}{2}$

Length of latus rectum $\frac{2 b^{2}}{a} = 6 \sqrt[]{2}$

$3 a = 6 \sqrt[]{2} \Rightarrow a = 2 \sqrt[]{2}$

$b = 2 \sqrt[]{3}$

y = 2x + c is tangent to hyperbola

$∴ c^{2} = a^{2} m^{2} - b^{2} = 20$

<math> <mrow> <msup> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mrow> <mo> (</mo> <mrow> <msup> <mrow> <mi>e</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math> <math> <mrow> <mi>e</mi> <mo>=</mo> <mroot> <mrow> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mrow></mrow> </mroot> <mo>⇒</mo> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <mn>3</mn> <msup> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math> Length of latus rectum <math> <mrow> <mfrac> <mrow> <mn>2</mn> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mi>a</mi> </mrow> </mfrac> <mo>=</mo> <mn>6</mn> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </math>   <math> <mrow> <mn>3</mn> <mi>a</mi> <mo>=</mo> <mn>6</mn> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> <mo>⇒</mo> <mi>a</mi> <mo>=</mo> <mn>2</mn> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </math>   <math> <mrow> <mi>b</mi> <mo>=</mo> <mn>2</mn> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> </math>  y = 2x + c is tangent to hyperbola <math> <mrow> <mo>∴</mo> <msup> <mrow> <mi>c</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <msup> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msup> <mrow> <mi>m</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mn>2</mn> <mn>0</mn> </mrow> </math>

0 Upvotes 0 Downvotes

Similar Questions for you

An ellipse whose length of minor axis is equal to half of length between foci, then eccentricity is

alok kumar singh

$?$ ae = 2b

$\frac{4 b^{2}}{a^{2}} = e^{2}$

Or 4 (1 – e²) = e²

4 = 5e² -> $e = \frac{2}{\sqrt[]{5}}$

The range of r for which circles (x + 1)² + (y + 2)² = r² and x² + y² – 4x – 4y + 4 = 0 coincide at two distinct points

alok kumar singh

If two circles intersect at two distinct points

->|r₁ – r₂| < C₁C₂ < r₁ + r₂

| r – 2| < $\sqrt[]{9 + 16}$ < r + 2

|r – 2| < 5 and r + 2 > 5

–5 < r 2 < 5 r > 3 … (2)

–3 < r < 7 (1)

From (1) and (2)

3 < r < 7

If hyperbola and ellipse x² – y²cosec²q = 5 and ellipse x²cosec²q + y² = 5 has eccentricity e_H and e_e respectively and e_H = $\sqrt[]{7} e_{e}$ , then q is equal to

alok kumar singh

x² – y²cosec²q = 5 $\frac{x^{2}}{1} - \frac{y^{2}}{s i n^{2} θ} = 5$

x² cosec²q + y² = 5 $\frac{x^{2}}{s i n^{2} θ} + \frac{y^{2}}{1} = 5$

$e_{H} = \sqrt[]{7} e_{e}$

and $e_{H} = \sqrt[]{1 + \frac{s i n^{2} θ}{1}}$

-> $\sqrt[]{1 + s i n^{2} θ} = \sqrt[]{7} \sqrt[]{1 - s i n^{2} θ}$

1 + sin²q = 7 – 7 sin²q

->8sin²q = 6

-> $s i n θ = \sqrt[]{\frac{3}{4}} = \frac{\sqrt[]{3}}{2}$

-> $θ = \frac{π}{3}$

A parabola has vertex (2, 3), equation of directrix is 2x – y = 1 and equation of ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, e = \frac{1}{\sqrt[]{2}}$ and ellipse passing through focur of parabola then square of length of latus rectum of ellipse is

alok kumar singh

Slope of axis = $\frac{1}{2}$

$y - 3 = \frac{1}{2} (x - 2)$

⇒ 2y – 6 = x – 2

⇒ 2y – x – 4 = 0

2x + y – 6 = 0

4x + 2y – 12 = 0

α + 1.6 = 4 ⇒ α = 2.4

β + 2.8 = 6 ⇒ β = 3.2

Ellipse passes through (2.4, 3.2)

⇒ $\frac{{(\frac{24}{10})}^{2}}{a^{2}} + \frac{{(\frac{32}{10})}^{2}}{b^{2}} = 1$

...more

Slope of axis = $\frac{1}{2}$

$y - 3 = \frac{1}{2} (x - 2)$

⇒ 2y – 6 = x – 2

⇒ 2y – x – 4 = 0

2x + y – 6 = 0

4x + 2y – 12 = 0

α + 1.6 = 4 ⇒ α = 2.4

β + 2.8 = 6 ⇒ β = 3.2

Ellipse passes through (2.4, 3.2)

⇒ $\frac{{(\frac{24}{10})}^{2}}{a^{2}} + \frac{{(\frac{32}{10})}^{2}}{b^{2}} = 1$

Also $1 - \frac{a^{2}}{b^{2}} = \frac{1}{2}$

$\frac{a^{2}}{b^{2}} = \frac{1}{2}$

$\frac{144}{25} b^{2} + \frac{256}{25} a^{2} = a^{2} b^{2}$

$\frac{144}{25} + \frac{256}{25} \times \frac{1}{2} = a^{2}$

$\frac{(128 + 144)}{25} = a^{2} \Rightarrow \frac{272}{25} = a^{2}$

-> $b^{2} = \frac{2 \times 272}{25}$

Latus rectum = $\frac{2 b^{2}}{a}$

(Latus rectum)²

$= \frac{4 b^{4}}{a^{2}} = 4 (\frac{b^{2}}{a^{2}}) b^{2} = \frac{8 \times 272 \times 2}{25} = \frac{4352}{25}$

less

The locus of the point of intersection of the lines $(\sqrt[]{3}) k x + k y - 4 \sqrt[]{3} = 0 a n d \sqrt[]{3} x - y - 4 (\sqrt[]{3}) k$ = 0 is a conic, whose eccentricity is…………………

Vishal Baghel

K $= \frac{4 \sqrt[]{3}}{\sqrt[]{3} x + y}$ $\frac{}{}$

K = $\frac{\sqrt[]{3} x - y}{4 \sqrt[]{3}}$ $\frac{}{}$

$∴ \frac{4 \sqrt[]{3}}{\sqrt[]{3} x + y} = \frac{\sqrt[]{3} x - y}{4 \sqrt[]{3}}$

$\frac{x^{2}}{16} - \frac{y^{2}}{48} = 1$

48 = 16 (e2 – 1)

$e = \sqrt[]{4} = 2$

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Let the eccentricity of the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 b e \sqrt[]{\frac{5}{2}}$ and length of its latus rectum be $6 \sqrt[]{2}, i f y = 2 x + c$ is a tangent to the hyperbola H, then the value of c² is equal to

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

Let the eccentricity of the hyperbola H : x 2 a 2 − y 2 b 2 = 1 b e 5 2 and length of its latus rectum be 6 2 , i f y = 2 x + c is a tangent to the hyperbola H, then the value of c2 is equal to

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

Let the eccentricity of the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 b e \sqrt[]{\frac{5}{2}}$ and length of its latus rectum be $6 \sqrt[]{2}, i f y = 2 x + c$ is a tangent to the hyperbola H, then the value of c² is equal to