A point on the ellipse $4{\mathrm{x}}^2 + 9{\mathrm{y}}^2 = 36$ where the normal is parallel to the line $4\mathrm{x} – 2\mathrm{y} – 5= 0$ is

$$\begin{gathered} \mathrm{eq} \mathrm{of} \mathrm{an} \mathrm{ellipse} \\ \frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{4}=1 \\ Let P(3\cos \theta ,2\sin \theta ) be any point on ellipse \\ i.e \mathrm{x}=3\mathrm{cos\theta } \mathrm{y}=2\mathrm{sin\theta } \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{cos\theta }}{-3 \mathrm{sin\theta }} \\ \mathrm{normal} \mathrm{slope}=\frac{3\mathrm{sin\theta }}{2\mathrm{cos\theta }}=\frac{4}{2} \end{gathered}$$

$\mathrm{Tan\theta }=\frac{4}{3}$  

$$\begin{gathered} \mathrm{point} \left(3.\frac{3}{5}, 2.\frac{4}{5}\right) \\ =\left(\frac{9}{5}, \frac{8}{5}\right) \end{gathered}$$