Question

Find normal to the ellipse $\frac{x^2}{12}+\frac{y^2}{8}=1$ at $(3,\sqrt{2})$

Answer 1

Equation of ellipse $\frac{x^2}{12}+\frac{y^2}{8}=1$

The slope of tangent $\frac{2x}{12}+\frac{2y}{8}\frac{dy}{dx}=0\frac{dy}{dx}=\frac{-2x}{3y}{\frac{dy}{dx}|}_{(3,\sqrt{2})}=-\sqrt{2}$

The slope of normal is $\frac{1}{\sqrt{2}}$

The equation of normal is $y-\sqrt{2}=\frac{1}{\sqrt{2}}(x-3)\sqrt{2}y-2=x-3x-\sqrt{2}y=1$