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 $\left(\pm4,0\right)$.
Let equation of required hyperbola be $\frac{x^{2}}{a'^{2}}-\frac{y^{2}}{b'^{2}}=1$
Let $e'$ be the eccentricity of required hyperbola then
$a'e'=4$
$\Rightarrow 2a'=4$ [$\because e'=2$ given]
$\Rightarrow a'=2$
$\therefore b'^{2}=a'^{2}\left(e'^{2}-1\right)$
$\Rightarrow b'^{2}=4\left(4-1\right)=12$
Thus equation of required hyperbola is $ \frac{x^{2}}{4}-\frac{y^{2}}{12}=1$.