If a tangent to the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1(\mathrm{a}>\mathrm{b})$ meets its major axis and minor axis at M and N respectively then prove that $\frac{{\mathrm{a}}^2}{(\mathrm{CM}{)}^2}+\frac{{\mathrm{b}}^2}{(\mathrm{CN}{)}^2}=1$ where C is the centre of the ellipse.

Let $\mathrm{p}\left(\mathrm{\theta }\right)=(\mathrm{asin\theta },\mathrm{bsin\theta })$ be a point on the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$.
Then the equation of the tangent at p($\mathrm{\theta }$) is $\frac{\mathrm{xcos\theta }}{\mathrm{a}}+\frac{\mathrm{ysin\theta }}{\mathrm{b}}=1$ 
$\text{ i.e., }\frac{\mathrm{x}}{\left(\frac{\mathrm{a}}{\cos \mathrm{\theta }}\right)}+\frac{\mathrm{y}}{\left(\frac{\mathrm{b}}{\sin \mathrm{\theta }}\right)}=1$
this line meets major axis (x-axis) and minor axis (y-axis) at M and N respectively
$\begin{array}{l}\therefore \mathrm{CM}=\frac{\mathrm{a}}{\cos \mathrm{\theta }};\mathrm{CN}=\frac{\mathrm{b}}{\sin \mathrm{\theta }}\\ \Rightarrow \cos \mathrm{\theta }=\frac{a}{\mathrm{CM}}\Rightarrow \left(1\right);\sin \mathrm{\theta }=\frac{\mathrm{b}}{\mathrm{CN}}\to \left(2\right)\\ (1{)}^2+(2{)}^2\Rightarrow {\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }=\frac{{\mathrm{a}}^2}{(\mathrm{CM}{)}^2}+\frac{{\mathrm{b}}^2}{(\mathrm{CN}{)}^2}\\ \Rightarrow 1=\frac{{\mathrm{a}}^2}{(\mathrm{CM}{)}^2}+\frac{{\mathrm{b}}^2}{(\mathrm{CN}{)}^2}\end{array}$