Question

An ellipse passes through the foci of the hyperbola, $9x^2—4y^2=36$ an 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. $\left(\sqrt{13},0\right)$
• B. $\left(\frac{\sqrt{39}}{2},\sqrt{3}\right)$
• C. $\left(\frac{1}{2},\sqrt{13},\frac{\sqrt{3}}{2}\right)$
• D. $\left(\sqrt{\frac{13}{2}},\sqrt{6}\right)$

Answer 1

Answer: (C)

Equation of hyperbola is
$\frac{x^2}{4}-\frac{y^2}{9}=1$
Its Foci $=\left(\pm \sqrt{13},0\right)$
$e=\frac{\sqrt{13}}{2}$
If e, be 2 the eccentricity of the ellipse, then
$e_1\times \frac{\sqrt{13}}{2}=\frac{1}{2}$
$\Rightarrow e_1=\frac{1}{\sqrt{13}}$
Equation of ellipse is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Since ellipse passes through the foci
$\left(\pm \sqrt{13},0\right)$ of the hyperbola, therefore
$a^2=13$
Now $\sqrt{a^2-b^2}=a{e}_1$
$\therefore 13-b^2=1$
$\Rightarrow b^{12}=12$
Hence, equation of ellipse is
$\frac{x^2}{13}+\frac{y^2}{12}=1$
Now putting the coordinate of the point
$\left(\frac{\sqrt{13}}{2},\frac{\sqrt{3}}{2}\right)$ in the equation of the ellipse,
we get
$\frac{13}{4\times 13}+\frac{3}{4\times 12}=1$
$\Rightarrow \frac{1}{4}+\frac{1}{16}=1$, which is not true,
Hence the point $\left(\frac{\sqrt{13}}{2},\frac{\sqrt{3}}{2}\right)$ does not lie on the ellipse.