Eccentric Ellipse

From calculus, we know that the tangent at $P(\phi)$ is:

$\dfrac{x\cos \phi}{a} + \dfrac{y \sin \phi}{b} = 1$

This line intersects the ellipse at two points $Q$ and $R$ . We can obtain the combined equation of the pair of lines $OQ$ and $OR$ by homogenizing the auxiliary circle's equation using the line equation.

The auxiliary circle's equation is given by:

$\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2} = 1$

Homogenizing:

$\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2} = \left(\dfrac{x\cos \phi}{a} + \dfrac{y \sin \phi}{b}\right)^2$

$\Longrightarrow \left(\dfrac{\sin^2 \phi}{a^2} \right)x^2 + \left(\dfrac{1}{a^2} -\dfrac{\sin^2 \theta}{b^2}\right)y^2 - \left(\dfrac{\sin 2\theta}{ab}\right)xy=0$

This pair of straight lines represents $OQ$ and $OR$ . We know that the angle between the pair of lines $ax^2+2hxy+by^2=0$ is given by:

$\theta = \tan^{-1} \left(\dfrac{\sqrt{h^2-ab}}{a+b}\right)$ .

For $\theta = \dfrac{\pi}{2}$ , $a+b=0$ :

$\Longrightarrow \dfrac{\sin^2 \phi}{a^2} + \dfrac{1}{a^2} -\dfrac{\sin^2 \phi}{b^2} = 0$

Using the relation $b^2=a^2(1-e^2)$ , we can eliminate $a$ and $b$ from the above to obtain:

$e\sqrt{1+sin^2 \phi}=\boxed{1}$