Question

Find the equation of a hyperbola with eccentricity $2$ and whose foci coincide with the ellipse $x^2/25+y^2/9=1$.

Answer 1

The equation of the ellipse is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a^2=25$ and $b^2=9$

Let $e$ be the eccentricity of the ellipse,

Then,

$\Rightarrow e=\sqrt{\frac{1-b^2}{a^2}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$

so the coordinates of foci are $(\pm ae,0)$ i.e, $(\pm 4,0)$

It is given that the foci of the hyperbola coincide with foci of ellipse. So the coordinates of foci of the hyperbola are $(\pm 4,0)$

Let $e$ be the eccentricity of the required hyperbola and its equation be -

$\Rightarrow \frac{x^2}{a^{\prime 2}}=\frac{-y^2}{b^{\prime 2}}=1$

Coordinates of the foci are $(\pm a^{\prime }{e}^{\prime },0)$

$\therefore a^{\prime }{e}^{\prime }=4$

$\Rightarrow 2a^{\prime }=4$

$\Rightarrow a^{\prime }=2$

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

$=2^2(4-1)$

$=4\times 3$

$=12$

substituting values of $a^{\prime }$ and $b^{\prime }$ in $\left(1\right)$

$\frac{x^2}{4}-\frac{y^2}{12}=1$ is equation of hyperbola.

Hence, solved.