Explanation

(c)
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 perpendicular from the focus $(ae,0)$ on the ellipse $= p$
$= \left| \frac{e\cos\theta - 1}{\sqrt{\frac{\cos^{2}\theta}{a^{2}} + \frac{\sin^{2}\theta}{b^{2}}}} \right|$
$= \left| \frac{ab(e\cos{\theta - 1)}}{\sqrt{b^{2}\cos^{2}{\theta +}a^{2}{({1 - cos}^{2}}{\theta)}}} \right| = \left| \frac{ab(e\cos{\theta - 1)}}{\sqrt{a^{2} - a^{2}e^{2}\cos^{2}\theta}} \right|$
$\Rightarrow b\sqrt{\frac{1 - e\cos\theta}{1 + e\cos\theta}} = p$
$\Rightarrow \frac{b^{2}}{p^{2}} = \frac{1 + e\cos\theta}{1 - e\cos\theta}$
Now, $r^{2} = {(ae - a\cos{\theta)}}^{2} + b^{2}\sin^{2}\theta$
$= a^{2}\left\lbrack {(e{- \cos}{\theta)}}^{2} + (1 - e)^{2}\sin^{2}\theta \right\rbrack$
$= a^{2}\left\lbrack e^{2}\cos^{2}{\theta - 2e\cos\theta + 1} \right\rbrack = a^{2}(1 - {e\cos{\theta)}}^{2}$
$\Rightarrow r = a(1 - e\ \cos{\theta)}$
$\therefore\frac{2a}{r} - \frac{b^{2}}{p^{2}} = \frac{2}{1 - e\cos\theta} - \frac{1 + e\cos\theta}{1 - e\cos\theta} = 1$