Question

The eccentric angles of extremities of a focal chord (other than Major axis) of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are ${\theta }_1$ and ${\theta }_2$.
If the eccentricity of the ellipse are $e_1$ and $e_2$ for the conditions $a>b$ and $b>a$ respectively, then ${\cos }^2\left(\frac{{\theta }_1-{\theta }_2}{2}\right)(\frac{1}{e_1^2}+\frac{1}{e_2^2})$ is

Answer 1

Let the coordinates of end points of the chord be
$P(a\cos {\theta }_1,b\sin {\theta }_1)$ and $Q(a\cos {\theta }_2,b\sin {\theta }_2)$
Equation the Focal chord is,
$\frac{y-b\sin {\theta }_2}{x-a\cos {\theta }_2}=\frac{b\sin {\theta }_1-b\sin {\theta }_2}{a\cos {\theta }_1-a\cos {\theta }_2}\Rightarrow \frac{y-b\sin {\theta }_2}{x-a\cos {\theta }_2}=-\frac{b}{a}\frac{\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}{\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}\cdots \left(1\right)$

When $a>b$,
Focus will be $(a{e}_1,0)$, putting this point in the equation of the focal chord,
$\frac{-b\sin {\theta }_2}{a{e}_1-a\cos {\theta }_2}=-\frac{b}{a}\frac{\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}{\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}\Rightarrow e_1\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)=\cos {\theta }_2\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)+\sin {\theta }_2\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)\Rightarrow e_1=\frac{\cos \left(\frac{{\theta }_1-{\theta }_2}{2}\right)}{\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}$

When $a<b$,
Focus will be $(0,b{e}_2)$, putting this point in the equation of the focal chord,
$\Rightarrow \frac{b{e}_2-b\sin {\theta }_2}{-a\cos {\theta }_2}=-\frac{b}{a}\frac{\cos \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}{\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}\Rightarrow e_2=\frac{\cos \left(\frac{{\theta }_1-{\theta }_2}{2}\right)}{\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)}$

Therefore,
$\frac{1}{e_1^2}+\frac{1}{e_2^2}=\frac{1}{{\cos }^2\left(\frac{{\theta }_1-{\theta }_2}{2}\right)}\Rightarrow {\cos }^2\left(\frac{{\theta }_1-{\theta }_2}{2}\right)(\frac{1}{e_1^2}+\frac{1}{e_2^2})=1$