The equation of any normal to $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is
$a x \sec \phi-b y \operatorname{cosec} \phi=a^{2}-b^{2} \,\,\,\,\,\,...(i)$
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 \,\,\, \frac{a \sec \phi}{\cos \alpha}=\frac{-b {cosec} \phi}{\sin \alpha}=\frac{a^{2}-b^{2}}{p} $
$\Rightarrow \,\,\, \cos \phi=\frac{a p}{\left(a^{2}-b^{2}\right) \cos \alpha}$
$ \sin \phi=\frac{-b p}{\left(a^{2}-b^{2}\right) \sin \alpha} $
$\because \,\,\,\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} {cosec}^{2} \alpha+a^{2} \sec ^{2} \alpha\right)=\left(a^{2}-b^{2}\right)^{2} $