Q.Find the equation of the tangent and normal to the ellipse $9x^2+16y^2=144$ at the end of the latus rectum in the first quadrant.

Find the latus-rectum end in the first quadrant, then apply the standard tangent and normal formulas for an ellipse at a point.

Divide by $144$: $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$, so $a^2=16$ ($a=4$), $b^2=9$ ($b=3$).

$c^2=a^2-b^2=16-9=7 \Rightarrow c=\sqrt7$

The end of the latus rectum in the first quadrant is $\left(\sqrt7,\dfrac{b^2}{a}\right)=\left(\sqrt7,\dfrac94\right)$

Tangent at $(x_1,y_1)$: $\dfrac{xx_1}{a^2}+\dfrac{yy_1}{b^2}=1$

$\dfrac{x\sqrt7}{16}+\dfrac{y(9/4)}{9}=1 \Rightarrow \dfrac{\sqrt7 x}{16}+\dfrac{y}{4}=1$

Multiply by 16: $\sqrt7\,x+4y=16$

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