Question

Let the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be the reciprocal to that of the ellipse $x^2+4y^2=4$. If the hyperbola passes through the focus of the ellipse, then the focus of the hyperbola is at :

• A. $(2,0)$
• B. $(0,2)$
• C. $(3,0)$
• D. $(0,3)$

Answer 1

The correct option is A $(2,0)$

$x^2+4y^2=4\Rightarrow \frac{x^2}{4}+y^2=1$

Eccentricity of ellipse $=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}=e$

Eccentricity of hyperbola $=\frac{1}{e}=\frac{2}{\sqrt{3}}=\sqrt{1+\frac{b^2}{a^2}}$
$\Rightarrow \frac{b}{a}=\frac{1}{\sqrt{3}}$

Given that the hyperbola passes through focus of ellipse.
Focus of ellipse $=(\pm ae,0)=(\pm \sqrt{3},0)$
Now in hyperbola equation, $\frac{3}{a^2}+0=1\Rightarrow a=\sqrt{3}$

Focus of hyperbola $=(\pm ae,0)=(\pm 2,0)$