The equation of a normal to the curve \(x=4 \sec \theta\) and \(y=4 \tan ^{2} \theta\) at \(\theta=\frac{\pi}{4}\) is

\(\begin{array}{rll} & x=4 \sec \theta & \text { and } & y=4 \tan ^{2} \theta \\ \therefore & \frac{\mathrm{dx}}{\mathrm{d} \theta}=4 \sec \theta \tan \theta & \text { and } & \frac{\mathrm{dy}}{\mathrm{d} \theta}=8 \tan \theta \cdot \sec ^{2} \theta\end{array}\)
\(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\) slope of tangent \(=\frac{8 \tan \theta \sec ^{2} \theta}{4 \sec \theta \tan \theta}=2 \sec \theta\)
At \(\theta=\frac{\pi}{4}, \quad \frac{\mathrm{dy}}{\mathrm{dx}}=2 \sqrt{2} \Rightarrow\) slope of normal \(=-\frac{1}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}=\frac{-1}{2 \sqrt{2}}\)
At \(\theta=\frac{\pi}{4}, \mathrm{x}=4 \sqrt{2}\) and \(\mathrm{y}=4\)
Hence eq. of normal is
\(\begin{aligned} &(y-4)=\frac{-1}{2 \sqrt{2}}(x-4 \sqrt{2}) \\ \therefore \quad & 2 \sqrt{2}(y-4)=-x+4 \sqrt{2} \Rightarrow x+2 \sqrt{2} y=12 \sqrt{2} \end{aligned}\)