Hyperbola Ques 13
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) $2$
(c) $\frac{13}{8}$
(d) $\frac{13}{6}$
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Answer:
Correct Answer: 13.(a)
Solution:
- We know that in $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, where $b^{2}=a^{2}\left(e^{2}-1\right)$, the length of conjugate axis is $2 b$ and distance between the foci is 2 ae.
$\therefore$ According the problem, $2 b=5$ and $2 a e=13$
Now,
$ b^{2}=a^{2}\left(e^{2}-1\right) $
$\Rightarrow \quad (\frac{5}{2}){ }^{2}=a^{2} e^{2}-a^{2}$
$\Rightarrow \quad \frac{25}{4}=\frac{(2 a e)^{2}}{4}-a^{2}$
$\Rightarrow \quad \frac{25}{4}=\frac{169}{4}-a^{2}$
$[\because 2 a e=13]$
$\Rightarrow \quad a^{2}=\frac{169-25}{4}=\frac{144}{4}=36$
$\Rightarrow \quad a=6$
Now, $\quad 2 a e=13$
$\Rightarrow \quad 2 \times 6 \times e=13$
$ \Rightarrow \quad e=\frac{13}{12} $
BETA
