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 $x^2+9y^2=9$, then the ratio $a^2:b^2$ equals

• A. 8 : 1
• B. 1 : 8
• C. 9 : 1
• D. 1 : 9

Answer 1

Answer: (A)

Given equation of ellipses is $x^2+9y^2=9$
$\Rightarrow \frac{x^2}{9}+\frac{y^2}{1}=1$
Here, $a=3,b=1$
$c=\sqrt{(3{)}^2-(1{)}^2}=\sqrt{8}$
$\therefore$ Eccentricity of ellipse, $e=\frac{c}{a}$
$\Rightarrow e=\frac{\sqrt{8}}{3}$
$\therefore$ Eccentricity of hyperbola $=\frac{3}{\sqrt{8}}$
$\Rightarrow 1+\frac{b^2}{a^2}=\frac{9}{8}$
$\Rightarrow \frac{b^2}{a^2}=\frac{1}{8}$
$\Rightarrow a^2:b^2=8:1$