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For an ellipse $\mathrm{ \frac{x^2}{9}+\frac{y^2}{4}=1}$ with vertices $\mathrm{ A}$ and $\mathrm{ A^{\prime}}$ , tangent drawn at the point $\mathrm{P}$ in the first quadrant meets the $\mathrm{y-axis}$ in $\mathrm{\mathrm{Q}}$ and the chord $\mathrm{\mathrm{A} ' \mathrm{P}}$ meets the $\mathrm{\mathrm{y}-axis }$ in $\mathrm{\mathrm{M}.}$ If $\mathrm{ ' \mathrm{O}'}$ is the origin then $\mathrm{ \mathrm{OQ}^2-\mathrm{MQ}^2}$ equals to

Option: 1
9

Option: 2
13

Option: 3
4

Option: 4
5

Answers (1)

best_answer

$\mathrm{a=3 ; b=2}$

$\mathrm{\mathrm{T}: \frac{\mathrm{x} \cos \theta}{3}+\frac{\mathrm{y} \sin \theta}{2}=1}$

$\mathrm{\mathrm{x}=0 ; \mathrm{y}=2 \operatorname{cosec} \theta}$

$\mathrm{chord \, \, A'P, \, \, \, \, \, \, y=\frac{2 \sin \theta}{3(\cos \theta+1)}(x+3) }$

$\mathrm{\text { put } x=0 \quad y=\frac{2 \sin \theta}{1+\cos \theta}=O M}$

$\mathrm{\text { Now } \mathrm{OQ}^2-\mathrm{MQ}^2=\mathrm{OQ}^2-(\mathrm{OQ}-\mathrm{OM})^2=2(\mathrm{OQ})(\mathrm{OM})-\mathrm{OM}^2=\mathrm{OM}\{2(\mathrm{OQ})-(\mathrm{OM})\}}$

$\mathrm{=\frac{2 \sin \theta}{1+\cos \theta}\left[\frac{y}{\sin \theta}-\frac{2 \sin \theta}{1+\cos \theta}\right]=4}$

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