The equation of any normal to $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $a x \sec \phi-b y \operatorname{cosec} \phi=a^{2}-b^{2} \quad \ldots$ (i)
The straight line $x \cos \alpha+y \sin \alpha=p$ will be a normal to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, if Eq. (i) and $x \cos \alpha+y \sin \alpha=p$ represent the same line.
$\therefore \quad \frac{a \sec \phi}{\cos \alpha}=\frac{-b \operatorname{cosec} \phi}{\sin \alpha}=\frac{a^{2}-b^{2}}{p}$
$\Rightarrow \quad \cos \phi=\frac{a p}{\left(a^{2}-b^{2}\right) \cos \alpha}$,
$\sin \phi=\frac{-b p}{\left(a^{2}-b^{2}\right) \sin \alpha}$
$\because \quad \sin ^{2} \phi+\cos ^{2} \phi=1$
$\Rightarrow \frac{b^{2} p^{2}}{\left(a^{2}-b^{2}\right)^{2} \sin ^{2} \alpha}+\frac{a^{2} p^{2}}{\left(a^{2}-b^{2}\right)^{2} \cos ^{2} \alpha}=1$
$\Rightarrow p^{2}\left(b^{2} \operatorname{cosec}^{2} \alpha+a^{2} \sec ^{2} \alpha\right)=\left(a^{2}-b^{2}\right)^{2}$