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

Let $p=\left(a\cos \theta ,b\sin \theta \right)$ since the eccentric angler of the extrimition of a pair of conjugate diameter differ by a right angle   $D=\left(a\cos \left(\frac{\pi }{2}+\theta \right),b\sin \left(\frac{\pi }{2}+\theta \right)\right)$

$=\left(-a\sin \theta ,b\cos \theta \right)$

$\therefore C{P}^2=a^2{\cos }^2\theta +b^2{\sin }^2\theta$

$C{D}^2=a^2{\sin }^2\theta +b^2{\cos }^2\theta$

$\therefore c{p}^2+c{d}^2=a^2+b^2$