For an ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ with vertices $A\left(3,0\right)$ $A\text{'}\left(-3,0\right)$ , tangent drawn at the point ‘$P$’ in the first quadrant meets the $y-$axis in $Q$ and the chord $A\text{'}P$ meets the $y-$axis in $M$. If ‘$O$’ is the centre of ellipse, then ${\mathrm{OQ}}^2-{\mathrm{MQ}}^2$ is equal to,

$\begin{array}{r}\mathrm{T}=\frac{\mathrm{xcos}\theta }{3}+\frac{\mathrm{ysin}\theta }{2}=1\\ \mathrm{x}=0,\mathrm{y}=2\mathrm{cosec}\theta \end{array}$
Chord A’P;
$\begin{array}{r}y=\frac{2\sin \theta }{3(\cos \theta +1)}(x+3)\\ x=0,OM=\frac{2\sin \theta }{(\cos \theta +1)}\end{array}$
Now, ${\mathrm{OQ}}^2-{\mathrm{MQ}}^2={\mathrm{OQ}}^2-(\mathrm{OQ}-\mathrm{OM}{)}^2$
$\begin{array}{l}=2\left(\mathrm{OM}\right)\left(\mathrm{OQ}\right)-{\mathrm{OM}}^2\\ =\mathrm{OM}[2\mathrm{OQ}-\mathrm{OM}]\\ =\frac{2\sin \theta }{(\cos \theta +1)}\left[\frac{4}{\sin \theta }-\frac{2\sin \theta }{1+\cos \theta }\right]\\ =\frac{2\sin \theta \left[4+4\cos \theta -2{\sin }^2\theta \right]}{(1+\cos \theta {)}^2\sin \theta }\\ \frac{4\sin \theta \left[2+2\cos \theta -1+{\cos }^2\theta \right]}{\left(1+\cos {\theta }^2\right)\sin \theta }\\ =4\end{array}$