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The equation of the hyperbola whose foci are the foci of the ellipse $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1$ and the eccentricity is $2$, is
$\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{12}} = 1$
$\frac{{{x^2}}}{4} - \frac{{{y^2}}}{{12}} = 1$
$\frac{{{x^2}}}{{12}} + \frac{{{y^2}}}{4} = 1$
$\frac{{{x^2}}}{{12}} - \frac{{{y^2}}}{4} = 1$
Solution
(b) Here for given ellipse $a = 5,\;b = 3,\;{b^2} = {a^2}(1 – {e^2})$
==> $e = \frac{4}{5}$
Therefore, focus is $(-4, 0), (4, 0).$
Given eccentricity of hyperbola = $2$
$a = \frac{{ae}}{e} = \frac{4}{2} = 2$ and $b = 2\sqrt {(4 – 1)} = 2\sqrt 3 $
Hence hyperbola is $\frac{{{x^2}}}{4} – \frac{{{y^2}}}{{12}} = 1$.
Similar Questions
Let $H : \frac{ x ^2}{ a ^2}-\frac{ y ^2}{ b ^2}=1$, where $a > b >0$, be $a$ hyperbola in the $xy$-plane whose conjugate axis $LM$ subtends an angle of $60^{\circ}$ at one of its vertices $N$. Let the area of the triangle $LMN$ be $4 \sqrt{3}$..
| List $I$ | List $II$ |
| $P$ The length of the conjugate axis of $H$ is | $1$ $8$ |
| $Q$ The eccentricity of $H$ is | $2$ ${\frac{4}{\sqrt{3}}}$ |
| $R$ The distance between the foci of $H$ is | $3$ ${\frac{2}{\sqrt{3}}}$ |
| $S$ The length of the latus rectum of $H$ is | $4$ $4$ |
The correct option is:
