Question

Let the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,(a>b)$ 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

• A. $(\pm 2,0)$
• B. $(\pm 2\sqrt{3},0)$
• C. $(\pm \frac{2}{\sqrt{3}},0)$
• D. $(\pm \frac{2}{3},0)$

Answer 1

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

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

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

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

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

Focus of hyperbola $=(\pm a{e}^{\prime },0)=(\pm 2,0)$