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If the product of focal distances of a point $\mathrm{P(a \cos \theta, b \sin \theta)}$ on an ellipse $\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$ is $\mathrm{\lambda}$ -times the square of the semi-diameter $\mathrm{CD}$ (of conjugate diameter $\mathrm{CP}$ ), then $\mathrm{ \lambda=}$

Option: 1
$4$

Option: 2
$2$

Option: 3
$1$

Option: 4
$3$

Answers (1)

best_answer

Let foci be $\mathrm{S(a e, 0) \: and \: S^{\prime}(-a e, 0)\: As\, P \: is \: (a \cos \theta, b \sin \theta)}$

$\mathrm{D \: is \: \: \left(a \cos \left(\theta+\frac{\pi}{2}\right), b \sin \left(\theta+\frac{\pi}{2}\right)\right)}$

$\mathrm{=(-a \sin \theta, b \cos \theta) . }$

By using definition of ellipse,

$\mathrm{ P S=e(P M) }$

$\mathrm{ \text { Or } \quad P S=e\left(\frac{a}{e}-a \cos \theta\right) }$

$\mathrm{ =a-a e \cos \theta }$

$\mathrm{ =a(1-e \cos \theta)(\text { Standard Result }) }$

$\mathrm{ \therefore \quad \text { SP.S'P }=a(1-e \cos \theta) a(1+e \cos \theta) }$

$\mathrm{ =a^2\left(1-e^2 \cos ^2 \theta\right)=a^2-a^2 e^2 \cos ^2 \theta }$

$\mathrm{ \text { But } b^2=a^2\left(1-e^2\right) }$

$\mathrm{ \therefore \quad \text { SP.S'P }=a^2-\left(a^2-b^2\right) \cos ^2 \theta }$

$\mathrm{ =a^2 \sin ^2 \theta+b^2 \cos ^2 \theta=\mathrm{CD}^2 }$

Hence option 3 is correct.

Posted by

jitender.kumar

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