Question:
Find the equation of the normal to the curve $x^{2}+2 y^{2}-4 x-6 y+8=0$ at the point whose abscissa is 2
Solution:
finding slope of the tangent by differentiating the curve
$2 x+4 y \frac{d y}{d x}-4-6 \frac{d y}{d x}=0$
$\frac{d y}{d x}=\frac{4-2 x}{4 y-6}$
Finding $y$ co - ordinate by substituting $x$ in the given curve
$2 y^{2}-6 y+4=0$
$y^{2}-3 y+2=0$
$y=2$ or $y=1$
$\mathrm{m}$ (tangent) at $\mathrm{x}=2$ is 0
normal is perpendicular to tangent so, $m_{1} m_{2}=-1$
$\mathrm{m}$ (normal) at $\mathrm{x}=2$ is $\frac{1}{0}$, which is undefined
equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$
$x=2$