The equation of the normal to the ellipse ${\mathrm{x}}^2+3{\mathrm{y}}^2=144$ at the positive end of the latusrectum is

Ellipse is $\frac{{\mathrm{x}}^2}{144}+\frac{{\mathrm{y}}^2}{48}=1\Rightarrow \mathrm{e}=\sqrt{\frac{144-48}{144}}=\sqrt{\frac{2}{3}}$
$\mathrm{L}=\left(\mathrm{ae},\frac{{\mathrm{b}}^2}{\mathrm{a}}\right)=\left(12\sqrt{\frac{2}{3}},4\right)$
$$\begin{gathered} =(4\sqrt{6},4)\text{ Normal is }\frac{a^{2}x}{x_1}\text{-}\frac{b^{2}y}{y_1}{\text{=a}}^2{\text{-b}}^2 \\ \text{⇒}\sqrt{3}\mathrm{x}-\sqrt{2}\mathrm{y}=8\sqrt{2} \end{gathered}$$