Question

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

• A. the equation of the hyperbola is $\frac{x^2}{3}-\frac{y^2}{2}=1$
• B. a focus of the hyperbola is $(2,0)$
• C. the eccentricity of the hyperbola is $\sqrt{\frac{5}{3}}$
• D. the equation of the hyperbola is $x^2-3y^2=3$

Answer 1

Answer: (B), (D)

a focus of the hyperbola is $(2,0)$
the equation of the hyperbola is $x^2-3y^2=3$

Given equation of ellipse can be written as $\frac{x^2}{4}+\frac{y^2}{1}=1$

So, the eccentricity $=\sqrt{1-\frac{b^2}{a^2}}$

$=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$

So, the eccentricity of hyperbola$=\frac{2}{\sqrt{3}}$

$\Rightarrow \sqrt{1+\frac{b^2}{a^2}}=\frac{2}{\sqrt{3}}$

$\Rightarrow 1+\frac{b^2}{a^2}=\frac{4}{3}$

$\Rightarrow \frac{b}{a}=\frac{1}{\sqrt{3}}$ ......(1)

Focus of ellipse $=(\pm ae,0)=(\pm \sqrt{3},0)$

So the hyperbola passes through $(\pm \sqrt{3},0)$

$\Rightarrow \frac{3}{a^2}=1$

or $a=\sqrt{3}$

So, by equation (1), we get $b=1$

So, the equation of hyperbola is

$\frac{x^2}{3}-\frac{y^2}{1}=1$

or, $x^2-3y^2=3$

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