If $P$ is the length of the perpendicular

 from a focus upon the tangent at any point P of the ellipse

$x^2/a^2+y^2/b^2=1$and $r$ r is the distance of P from the focus,  

then $\frac{2a}{r}-\frac{b^2}{p^2}=$ 

The equation of the tangent at $P(a\cos \theta$

$b\sin \theta )\mathrm{t}$   to the ellipse  $x^2/a^2+y^2/b^2=1$

 is$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$ 

length of the perpendicular from the focus$(ae, 0)$ on the line is 

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

Now, 

  $\begin{array}{l}{r}^2=(ae-a\cos \theta {)}^2+b^2{\sin }^2\theta \\ =a^2\left[(e-\cos \theta {)}^2+\left(1-e^2\right){\sin }^2\theta \right]\\ =a^2\left[e^2{\cos }^2\theta -2e\cos \theta +1\right]\\ =a^2(1-e\cos \theta {)}^2\\ \Rightarrow r=a(1-e\cos \theta )\end{array}$ 

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