Question

An ellipse passes through the foci of the hyperbola, $9x^2-4y^2=36$ and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively. If the product of eccentricities of the two conics is $\frac{1}{2}$, then which of the following points does not lie on the ellipse ?

• A. $(\sqrt{\frac{13}{2}},\sqrt{6})$
• B. $(\frac{1}{2}\sqrt{13},\frac{\sqrt{3}}{2})$
• C. $(\sqrt{13},0)$
• D. $(\frac{\sqrt{39}}{2},\sqrt{3})$

Answer 1

The correct option is B $(\frac{1}{2}\sqrt{13},\frac{\sqrt{3}}{2})$

Given equation of hyperbola is
$9x^2-4y^2=36$
$\Rightarrow \frac{x^2}{4}-\frac{y^2}{9}=1$
Here, $a=2,b=3$
$e^{\prime }=\sqrt{1+\frac{b^2}{a^2}}=\frac{\sqrt{13}}{2}$
Foci of hyperbola is $(\pm \sqrt{13},0)$
Given $e{e}^{\prime }=\frac{1}{2}$
So, eccentricity of ellipse is $\frac{1}{\sqrt{13}}$

ellipse passes through focus $(\pm \sqrt{13},0)$ of hyperbola

let ellipse is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

put point in above equation

we get $a^2=13$ , $b^2=12$

so equation of ellipse

$\frac{x^2}{13}+\frac{y^2}{12}=1$