A point on the ellipse ${\mathrm{x}}^2+3{\mathrm{y}}^2=37$ where the normal is parallel to the line 6x – 5y = 2, is

Equation of ellipse is $\frac{{\mathrm{x}}^2}{37}+\frac{{\mathrm{y}}^2}{37/3}=1$
$$\begin{gathered} where a^2=37,b^2=\frac{37}{3} \\ Let P(x_1,y_1) be any point on ellipse then \\ \frac{{x_1}^2}{37}+\frac{{y_1}^2}{{\displaystyle \frac{37}{3}}}=1\to \left(1\right) \\ Equation of normal at P is \\ \frac{37x}{x_1}-\frac{37y}{3y_1}=\frac{74}{2} \\ slope of normal=\frac{3y_1}{x_1} \\ Given it is parallel to the line 6x-5y-2=0 \\ then \frac{3y_1}{x_1}=\frac{6}{5} \\ \Rightarrow y_1=\frac{2}{5}{x}_1 \\ sub y_{1 } in \left(1\right) then \\ {x_1}^2+\frac{12}{25}{x_1}^2=37 \\ \Rightarrow {x_1}^2=25 \\ \Rightarrow x_1=\pm 5 \\ \Rightarrow y_1=\pm 2 \\ points are (5,2) and (-5,-2) \end{gathered}$$
$\therefore$point of contact = (5, 2) or (–5, –2)