Question

The foci of a hyperbola coincide with the fpci of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$. Find the equation of the hyperbola if its eccentricity is $2$.

• A. $\frac{x^2}{12}-\frac{y^2}{4}=1$
• B. $\frac{x^2}{4}-\frac{y^2}{12}=1$
• C. $\frac{x^2}{3}-\frac{y^2}{4}=1$
• D. $\frac{x^2}{4}-\frac{y^2}{3}=1$

Answer 1

Answer: (B)

The equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{9}=1$.
$\therefore a^2=25$ and $b^2=9$
Eccentricity of ellipse $=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$
So the co-ordinates of the foci are $\left(\pm 4,0\right)$.
$\therefore$ The foci of hyperbola are $\left(\pm 4,0\right)$.
Let equation of required hyperbola be $\frac{x^2}{a^{\prime 2}}-\frac{y^2}{b^{\prime 2}}=1$
Let $e^{\prime }$ be the eccentricity of required hyperbola then
$a^{\prime }{e}^{\prime }=4$
$\Rightarrow 2a^{\prime }=4$ [$\because e^{\prime }=2$ given]
$\Rightarrow a^{\prime }=2$
$\therefore b^{\prime 2}=a^{\prime 2}\left(e^{\prime 2}-1\right)$
$\Rightarrow b^{\prime 2}=4\left(4-1\right)=12$
Thus equation of required hyperbola is $\frac{x^2}{4}-\frac{y^2}{12}=1$.