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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 ?

Option 1)
$\left ( \sqrt{13},0 \right )$

Option 2)
$\left ( \frac{\sqrt{39}}{2}, \sqrt{3}\right )$

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

Option 4)
$\sqrt{\frac{13}{2}},\sqrt{6}$

Answers (1)

best_answer

As we learnt in

Eccentricity -

$e= \sqrt{1-\frac{b^{2}}{a^{2}}}$

- wherein

For the ellipse

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

Coordinates of foci -

$\pm ae,o$

- wherein

For the ellipse

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

$9x^{2}-4y^{2}=36$

$=> \frac{x^{2}}{4}-\frac{y^{2}}{9}=1$

$e=\sqrt{\frac{1+9}{4}}$ $=\frac{\sqrt{13}}{2}$

$e_{1}\times e _{2}=\frac{1}{2}$

$\Rightarrow e_{2}=\frac{1}{\sqrt{13}}$

Also $ae_{1}=2\times \frac{\sqrt{13}}{2}$ $=\sqrt{13}$

Hence ellipse passes through $\left (\pm\sqrt{13, 0} \right )$

$For\: ellipse\:\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1$

$A^{2}=13$

$Also, B^{2} =A^{2}\left ( 1-e_{2}^{2} \right )$

$B^{2} = 13\left ( 1-\frac{1}{13} \right )$ $=12$

$\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1$

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

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

doesn't lie on this ellipse

Option 1)

$\left ( \sqrt{13},0 \right )$

Incorrect option

Option 2)

$\left ( \frac{\sqrt{39}}{2}, \sqrt{3}\right )$

Incorrect option

Option 3)

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

Correct option

Option 4)

$\sqrt{\frac{13}{2}},\sqrt{6}$

Incorrect option

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prateek

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