If $\mathrm{CP}$ and $\mathrm{CD}$ is a pair of semi-conjugate diameters of the ellipse…
If $\mathrm{CP}$ and $\mathrm{CD}$ 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
$\mathrm{CP}^{2}+\mathrm{CD}^{2}=$
$\frac{a^{2}+b^{2}}{2}$
$a^{2}+b^{2}$
$a^{2}-b^{2}$
$\frac{a^{2}-b^{2}}{2}$
Solution
The coordinates of $\mathrm{P}$ and $\mathrm{D}$ are $(a \cos \theta, b \sin \theta)$ and $\left(a \cos \left(\frac{\pi}{2}+\theta\right), b \sin \left(\frac{\pi}{2}+\theta\right)\right)$ respectively.
$\therefore C P^{2}+C D^{2}=a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta$
$\Rightarrow \quad C P^{2}+C D^{2}=a^{2}+b^{2}$