If CP and CD is a pair of semi-conjugate diameters of the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ then CP²+CD²=

Let =P(acos$\mathrm{\theta }$, bsin$\mathrm{\theta }$) be  the eccentric angle of the extremity of a pair of conjugate diameter differ by a right angle.
$\begin{array}{l}\mathrm{D}=\left(\mathrm{acos}\left(\frac{\mathrm{\pi }}{2}+\mathrm{\theta }\right),\mathrm{bsin}\left(\frac{\mathrm{\pi }}{2}+\mathrm{\theta }\right)\right)\\ =(-\mathrm{asin}\mathrm{\theta },\mathrm{bcos}\mathrm{\theta })\\ \therefore {\mathrm{CP}}^2={\mathrm{a}}^2{\cos }^2\mathrm{\theta }+{\mathrm{b}}^2{\sin }^2\mathrm{\theta }\\ {\mathrm{CD}}^2={\mathrm{a}}^2{\sin }^2\mathrm{\theta }+{\mathrm{b}}^2{\cos }^2\mathrm{\theta }\\ \therefore C{P}^2+C{D}^2={\mathrm{a}}^2+{\mathrm{b}}^2\end{array}$