Question

If $p$ is the length of the perpendicular from focus upon the tangent at any point $P$ of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $r$ is the distance of $P$ from the focus, then $(\frac{2a}{r}-\frac{b^2}{p^2})$ is equal to

Answer 1

Equation of tangent at point $P$ of ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$
Given $p$ is the distance of tangent from $F_2(ae,0)$
Then, $p=\frac{|e\cos \theta -1|}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}=\frac{|e\cos \theta -1|}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{a^2(1-e^2)}}}=\frac{a|e\cos \theta -1|\sqrt{1-e^2}}{\sqrt{1-e^2{\cos }^2\theta }}$
$\frac{b^2}{p^2}=\frac{a^2(1-e^2)(1-e^2{\cos }^2\theta )}{a^2(1-e\cos \theta {)}^2(1-e^2)}=\frac{1+e\cos \theta }{1-e\cos \theta }\cdots \left(i\right)$
Also, $r$ is the distance of $P(a\cos \theta ,b\sin \theta )$ from $F_2(ae,0)$
$r=\sqrt{(a\cos \theta -ae{)}^2+b^2{\sin }^2\theta }=\sqrt{(a\cos \theta -ae{)}^2+a^2(1-e^2){\sin }^2\theta }=a(1-e\cos \theta )$
$\Rightarrow \frac{2a}{r}=\frac{2}{1-e\cos \theta }\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$\Rightarrow \frac{2a}{r}-\frac{b^2}{p^2}=\frac{2}{1-e\cos \theta }-\frac{1+e\cos \theta }{1-e\cos \theta }=1$