For an ellipse frac{{{x^2}}}{9} + frac{{{y^2}}}{4} = 1 with vertices A and A', tangent dr

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10-2. Parabola, Ellipse, Hyperbola

normal

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

$9$

$13$

$4$

$5$

Solution

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

Standard 11

Mathematics

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(KVPY-2014)

The locus of the point of intersection of perpendicular tangents to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, is

medium

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

Solution

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