Question

An ellipse passes through the foci of the hyperbola, $9x^2–4y^2=36$ and its major and minor axis lie along the transverse and conjugate axis 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. $(\sqrt{13},0)$
• C. $(\frac{1}{2}\sqrt{13},\frac{\sqrt{3}}{2})$
• D. $(\frac{\sqrt{39}}{2},\sqrt{3})$

Answer 1

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

$9x^2–4y^2=36$
$\Rightarrow \frac{x^2}{4}-\frac{y^2}{9}=1$
$e_1=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}$
$\therefore$ Coordinates of foci are $(\pm a{e}_1,0)\equiv (\pm \sqrt{13},0)$.

Let $e$ be the eccentricity of ellipse
Given, $e{e}_1=\frac{1}{2}$
$\therefore$ Eccentricity of ellipse, $e=\frac{1}{\sqrt{13}}$
Let equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$
Ellipse passes through the foci.
$\Rightarrow \frac{13}{a^2}=1\Rightarrow a^2=13$
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.