For an ellipse frac{{{x^2}}}{9} + frac{{{y^2} | vedclass

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Question

$a = 3 ; b = 2$
$T :\frac{{x\cos 	heta }}{3} + \frac{{y\sin 	heta }}{2} = 1$ 
$x = 0 ; y = 2 cosec 	heta$ 
chord $A'P,$ $y = \frac

Vedclass · Accepted Answer

$4$

Vedclass · Answer

$a = 3 ; b = 2$
$T :\frac{{x\cos 	heta }}{3} + \frac{{y\sin 	heta }}{2} = 1$ 
$x = 0 ; y = 2 cosec 	heta$ 
chord $A'P,$ $y = \frac{{2\sin 	heta }}{{3(\cos 	heta  + 1)}}(x + 3)$ 
put $x = 0$ $ y$  =$\frac{{2\sin 	heta }}{{1 + \cos 	heta }}$ = $OM$ 
Now $OQ^2 - MQ^2 = OQ^2 - (OQ - OM)^2$$ = 2(OQ)(OM) - OM^2 = OM\{ 2(OQ) - (OM) \} $ 
= $\frac{{2\sin 	heta }}{{1 + \cos 	heta }}$$\left[ {\frac{y}{{\sin 	heta }} - \frac{{2\sin 	heta }}{{1 + \cos 	heta }}} ight]$ $= 4$

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

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