The equation of the normal to the curve $2 x^{2}+y^{2}=12$ at the point $(2,2)$ is
The equation of the normal to the curve $2 x^{2}+y^{2}=12$ at the point $(2,2)$ is
$2 x-y+6=0$
$2 x+y-6=0$
$x+2 y+2=0$
$x-2 y+2=0$
Solution
Given equation of the curve is $2 x^{2}+y^{2}=12$
$\therefore \quad 4 x+2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{-2 x}{y}$
Slope of tangent at $(2,2)$ is $\frac{-2(2)}{2}-2 \Rightarrow$ Slope of normal at $(2,2)$ is $\frac{1}{2}$
Equation of normal at $(2,2)$ is
$y-2=\frac{1}{2}(x-2) \Rightarrow 2 y-4=x-2 \Rightarrow x-2 y+2=0$