Find the equation of tangent and normal to the ellipse 9x² + 16y² = 144 at the end of the latus rectum in the first quadrant.

Given Ellipse $9{\mathrm{x}}^2+16{\mathrm{y}}^2=144$
$\begin{array}{l}\frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{9}=1\Rightarrow \mathrm{a}=4,\mathrm{b}=3,\mathrm{a}>\mathrm{b}\\ \mathrm{e}=\sqrt{\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{{\mathrm{a}}^2}}=\sqrt{\frac{16-9}{16}}=\frac{\sqrt{7}}{4}\end{array}$
Positive end of L.R = $\left(\mathrm{ae},\frac{{\mathrm{b}}^2}{\mathrm{a}}\right)$
$=\left(4\cdot \frac{\sqrt{7}}{4},\frac{9}{4}\right)=\left(\sqrt{7},\frac{9}{4}\right)$
Equation of tangent to Ellipse $9{\mathrm{x}}^2+16{\mathrm{y}}^2=144$ is S₁ = 0
$9{\mathrm{xx}}_1+16{\mathrm{yy}}_1-144=0$
At point $=\left(\sqrt{7},\frac{9}{4}\right)\Rightarrow 9\mathrm{x}\cdot \sqrt{7}+16\mathrm{y}\cdot \frac{9}{4}-144=0$
$\Rightarrow \sqrt{7}\mathrm{x}+4\mathrm{y}-16=0$
Slope of tangent $=-\frac{\sqrt{7}}{4};\text{ Slope of normal }=\frac{4}{\sqrt{7}}$
$\text{ Equation of normal at }\left(\sqrt{7},\frac{9}{4}\right)\text{ is }$
$\begin{array}{l}\Rightarrow \mathrm{y}-\frac{9}{4}=\frac{4}{\sqrt{7}}(\mathrm{x}-\sqrt{7})\Rightarrow \mathrm{y}-\frac{9}{4}=\frac{4\mathrm{x}}{\sqrt{7}}-4\\ \Rightarrow \frac{4\mathrm{x}}{\sqrt{7}}-\mathrm{y}+\frac{9}{4}-4=0\\ \Rightarrow \frac{16\mathrm{x}-4\sqrt{7}\mathrm{y}+9\sqrt{7}-16\sqrt{7}}{4\sqrt{7}}=0\\ \Rightarrow 16\mathrm{x}-4\sqrt{7}\mathrm{y}-7\sqrt{7}=0\end{array}$