SATHEE: Hyperbola 1 Question 4

Hyperbola 1 Question 4

5.

If a hyperbola has length of its conjugate axis equal to $5$ and the distance between its foci is $13$ , then the eccentricity of the hyperbola is

(2019 Main, 11 Jan II)

(a) $\frac{13}{12}$

(b) $2$

(d) $\frac{13}{6}$

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Answer:

Correct Answer: 5. (a)

Solution:

We know that in $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , where $b^{2} = a^{2} (e^{2} - 1)$ , the length of conjugate axis is $2 b$ and distance between the foci is 2 ae.

$∴$ According the problem, $2 b = 5$ and $2 a e = 13$

Now,

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

$\Rightarrow \frac{5}{2}^{2} = a^{2} e^{2} - a^{2}$

$\Rightarrow \frac{25}{4} = \frac{(2 a e)^{2}}{4} - a^{2}$

$\Rightarrow \frac{25}{4} = \frac{169}{4} - a^{2}$

$[∵ 2 a e = 13]$

$\Rightarrow a^{2} = \frac{169 - 25}{4} = \frac{144}{4} = 36$

$\Rightarrow a = 6$

Now, $2 a e = 13$

$\Rightarrow 2 \times 6 \times e = 13$

$\Rightarrow e = \frac{13}{12}$

Hyperbola 1 Question 4

5.

Answer:

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Hyperbola 1 Question 4

5.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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