If $\mathrm{a}>\mathrm{b}$ and e is the eccentricity of the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ then the equation of the normal
at the end of the latusrectum in the first quadrant is

Given ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 (\mathrm{a}>\mathrm{b})$

one end of latusrectum $\left(\mathrm{L}\right)=\left(\mathrm{ae}, \frac{{\mathrm{b}}^2}{\mathrm{a}}\right)$ in ${\mathrm{Q}}_1$

Equation of normal at 'L' is 

$\begin{array}{rcl}\frac{{\mathrm{a}}^2\mathrm{x}}{\mathrm{ae}}-\frac{{\mathrm{b}}^2\mathrm{y}}{{\displaystyle \frac{{\mathrm{b}}^2}{\mathrm{a}}}}& =& {\mathrm{a}}^2-{\mathrm{b}}^2\\ & \Rightarrow & \frac{\mathrm{ax}}{\mathrm{e}}-\mathrm{ay}={\mathrm{a}}^2-{\mathrm{a}}^2(1-{\mathrm{e}}^2)\\ & \Rightarrow & \mathrm{a}\left[\frac{\mathrm{x}}{\mathrm{e}}-\mathrm{y}\right]={\mathrm{a}}^2\left(1-1+{\mathrm{e}}^2\right)\\ & \Rightarrow & \frac{\mathrm{x}-\mathrm{ey}}{\mathrm{e}}=\mathrm{a}{\left(\mathrm{e}\right)}^2 \Rightarrow \mathrm{x}-\mathrm{ey}={\mathrm{ae}}^3\\ & \Rightarrow & \mathrm{x}-\mathrm{ey}-{\mathrm{ae}}^3=0\end{array}$