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 $doesnot$ lie on the ellipse?

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

Answer 1

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

$9x^2–4y^2=36$
$\Rightarrow \frac{x^2}{4}-\frac{y^2}{9}=1$
$\therefore$ Coordinates of foci are $(\sqrt{13},0),(-\sqrt{13},0)$.
Eccentricity of hyperbola $=\frac{\sqrt{13}}{2}$
$\therefore$ Eccentricity of ellipse, $e=\frac{1}{\sqrt{13}}$
Let equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Ellipse passes through the foci.
$\Rightarrow \frac{13}{a^2}=1\Rightarrow a^2=13$
$\text{Also},e^2=1-\frac{b^2}{a^2}=\frac{1}{13}\Rightarrow b^2=12$
Thus, equation of ellipse is $\frac{x^2}{13}+\frac{y^2}{12}=1$
$\therefore (\frac{1}{2}\sqrt{13},\frac{\sqrt{3}}{2})$ does not lie on the ellipse.