For an ellipse frac{x^2}{9} + frac{y^2}{4} = | vedclass

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(C) Given the ellipse $\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$,we have $a = 3$ and $b = 2$.Let $P = (3 \cos 	heta, 2 \sin 	heta)$.The equation of the

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$4$

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(C) Given the ellipse $\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$,we have $a = 3$ and $b = 2$.Let $P = (3 \cos 	heta, 2 \sin 	heta)$.The equation of the tangent at $P$ is $\frac{x \cos 	heta}{3} + \frac{y \sin 	heta}{2} = 1$.For $Q$,set $x = 0$,so $y = \frac{2}{\sin 	heta} = 2 \csc 	heta$. Thus,$OQ = 2 \csc 	heta$.The coordinates of $A'$ are $(-3, 0)$. The equation of the chord $A'P$ is $y - 0 = \frac{2 \sin 	heta - 0}{3 \cos 	heta - (-3)} (x - (-3))$,which simplifies to $y = \frac{2 \sin 	heta}{3(1 + \cos 	heta)} (x + 3)$.For $M$,set $x = 0$,so $OM = \frac{2 \sin 	heta}{3(1 + \cos 	heta)} \cdot 3 = \frac{2 \sin 	heta}{1 + \cos 	heta} = 2 	an(	heta/2)$.We need $OQ^2 - MQ^2$. Note that $MQ = |OQ - OM| = OQ - OM$ (since $OQ > OM$ for $	heta \in (0, \pi/2)$).$OQ^2 - (OQ - OM)^2 = OQ^2 - (OQ^2 - 2(OQ)(OM) + OM^2) = 2(OQ)(OM) - OM^2$.Substitute $OQ = \frac{2}{\sin 	heta}$ and $OM = \frac{2 \sin 	heta}{1 + \cos 	heta} = \frac{2 \sin 	heta}{2 \cos^2(	heta/2)} = 2 	an(	heta/2)$.$2(OQ)(OM) - OM^2 = 2 \left( \frac{2}{\sin 	heta} ight) \left( \frac{2 \sin 	heta}{1 + \cos 	heta} ight) - \left( \frac{2 \sin 	heta}{1 + \cos 	heta} ight)^2 = \frac{8}{1 + \cos 	heta} - \frac{4 \sin^2 	heta}{(1 + \cos 	heta)^2} = \frac{8(1 + \cos 	heta) - 4(1 - \cos^2 	heta)}{(1 + \cos 	heta)^2} = \frac{8 + 8 \cos 	heta - 4 + 4 \cos^2 	heta}{(1 + \cos 	heta)^2} = \frac{4(1 + 2 \cos 	heta + \cos^2 	heta)}{(1 + \cos 	heta)^2} = \frac{4(1 + \cos 	heta)^2}{(1 + \cos 	heta)^2} = 4$.

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

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