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)

$\left(\frac{1}{2},\sqrt{13}, \frac{\sqrt{3}}{2}\right)$

Answer 2

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 $\sqrt{a^{2}-b^{2}} = ae_{1}$
$\therefore 13-b^{2}= 1$
$\Rightarrow b^{12} = 12$
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
$\frac{13}{4\times13}+\frac{3}{4\times 12} = 1$
$\Rightarrow \frac{1}{4}+\frac{1}{16} = 1$, which is not true,
Hence the point $\left(\frac{\sqrt{13}}{2},\frac{\sqrt{3}}{2}\right)$ does not lie on the ellipse.