If the line \(x \cos \alpha+y \sin \alpha=p\) be normal to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), then

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.
\(\therefore \quad \frac{a \sec \phi}{\cos \alpha}=\frac{-b \operatorname{cosec} \phi}{\sin \alpha}=\frac{a^{2}-b^{2}}{p}\)
\(\Rightarrow \quad \cos \phi=\frac{a p}{\left(a^{2}-b^{2}\right) \cos \alpha}\),
\(\sin \phi=\frac{-b p}{\left(a^{2}-b^{2}\right) \sin \alpha}\)
\(\because \quad \sin ^{2} \phi+\cos ^{2} \phi=1\)
\(\Rightarrow \frac{b^{2} p^{2}}{\left(a^{2}-b^{2}\right)^{2} \sin ^{2} \alpha}+\frac{a^{2} p^{2}}{\left(a^{2}-b^{2}\right)^{2} \cos ^{2} \alpha}=1\)
\(\Rightarrow p^{2}\left(b^{2} \operatorname{cosec}^{2} \alpha+a^{2} \sec ^{2} \alpha\right)=\left(a^{2}-b^{2}\right)^{2}\)