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

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10-2. Parabola, Ellipse, Hyperbola

hard

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

A

$\frac{{13}}{{12}}$

B

$2$

C

$\frac{{13}}{{6}}$

D

$\frac{{13}}{{8}}$

(JEE MAIN-2019)

Solution

$2b = 5$ and $2ae = 13$

${\left( {ae} \right)^2} = {a^2} + {b^2}$

$ \Rightarrow {a^2} = {\left( {ae} \right)^2} – {b^2} = \frac{{169}}{4} – \frac{{25}}{4}$

$ \Rightarrow a = 6$

$e = \frac{{ae}}{a} = \frac{{13}}{{12}}$

Standard 11

Mathematics

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