Question

An ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the hyperbola $2 x^2-2 y^2=-1$ intersect orthogonally. If product of eccentricity of the ellipse and hyperbola is 1 , then $\frac{a^2}{b^2}$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. 2
• D. 4

Answer 1

Answer: (A)

$\left(\frac{x_0}{y_0}\right)\left(\frac{-b^2 x_0}{a^2 y_0}\right)=-1$

$\frac{ a ^2}{ b ^2}=\frac{ x _0^2}{ y _0^2}=\frac{ x _0^2}{ x _0^2+\frac{1}{2}}<1$
eccentricity of ellipse
$e ^2=1-\frac{ a ^2}{ b ^2}=\frac{1}{2} \Rightarrow \frac{ a ^2}{ b ^2}=\frac{1}{2}$