Question

If $p$ is the length of the perpendicular from a 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 $\frac{2a}{r}-\frac{b^2}{p^2}=$

Answer 1

The equation of the tangent at $P(a\cos \theta ,b\sin \theta )$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1.$

Length of the perpendicular from the focus $(ae,0)$ on the ellipse

is $p=\left|\frac{e\cos \theta -1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}\right|=\left|\frac{ab(e\cos \theta -1)}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}\right|$

$=\left|\frac{ab(e\cos \theta -1)}{\sqrt{a^2-a^2{e}^2{\cos }^2\theta }}\right|=b\sqrt{\frac{1-e\cos \theta }{1+e\cos \theta }}$

$\Rightarrow \frac{b^2}{p^2}=\frac{1+e\cos \theta }{1-e\cos \theta }$

Now, $r^2={(ae-a\cos \theta )}^2+b^2{\sin }^2\theta =a^2[{(e-\cos \theta )}^2+(1-e^2){\sin }^2\theta ]$

$=a^2[e^2{\cos }^2\theta -2e\cos \theta +1]$

$=a^2{(1-e\cos \theta )}^2$

$\Rightarrow r=a(1-e\cos \theta )$

Now, $\frac{2a}{r}-\frac{b^2}{p^2}=\frac{2}{1-e\cos \theta }-\frac{1+e\cos \theta }{1-e\cos \theta }=1$