If the normals at $\mathrm{P}\left(\mathrm{\theta }\right)\text{ and }\mathrm{Q}(\mathrm{\pi }/2+\mathrm{\theta })$ to the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ meet the major axis at G and g, respectively, then ${\mathrm{PG}}^2+{\mathrm{Qg}}^2=$

$Given ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Normal at$P\left(\theta \right) is \frac{\mathrm{ax}}{\cos \mathrm{\theta }}-\frac{\mathrm{by}}{\sin \mathrm{\theta }}={\mathrm{a}}^2{\mathrm{e}}^2$  

Equation of normal at $\mathrm{Q}\left(\frac{\mathrm{\pi }}{2}+\mathrm{\theta }\right)\text{ is }\frac{\mathrm{ax}}{-\sin \mathrm{\theta }}-\frac{\mathrm{by}}{\cos \mathrm{\theta }}={\mathrm{a}}^2{\mathrm{e}}^2$

Given both  normals cut major axis at G and g respectively then 
$\begin{array}{l}\mathrm{G}\left({\mathrm{ae}}^2\cos \mathrm{\theta },0\right);\mathrm{g}=\left(-{\mathrm{ae}}^2\sin \mathrm{\theta },0\right)\\ \Rightarrow {\mathrm{PG}}^2+{\mathrm{Qg}}^2={\mathrm{a}}^2\left(1-{\mathrm{e}}^2\right)\left(2-{\mathrm{e}}^2\right)\end{array}$