Explanation:
A Given equation of parabola, \(y^2=16 x\)
Differentiating both sides, we get
\(2 y \frac{d y}{d x}=16\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{8}{\mathrm{y}}\)
\(\therefore\) Slope of at point \(\left(\mathrm{x}_1, \mathrm{y}_1\right), \mathrm{m}_1=\frac{8}{\mathrm{y}_1}\)
and also slope of normal at point \(\left(\mathrm{x}_1, \mathrm{y}_1\right), \mathrm{m}_2=\frac{-\mathrm{y}_1}{8}\)
Since, normal makes equal angle with both \(\mathrm{X}\) and \(\mathrm{Y}\) axes, then
\(\mathrm{m}_2= \pm 1\)
\(\frac{-\mathrm{y}_1}{8}= \pm 1\)
\(-\mathrm{y}_1= \pm 8\)
When \(\mathrm{y}_1=8\), then \(\mathrm{x}_1=4\)
When \(\mathrm{y}_1=-8\) then \(\mathrm{x}_1=4\)
Hence, the required point is \((4,-8)\).