Question

Find the equation of the normal to curve $x^2=4y$ which passes through the point $(1,2)$.

Answer 1

Slope of tangent to the curve will be
$\frac{dy}{dx}=\frac{x}{2}$
Slope of normal will be $\frac{-2}{x}$
Point is given by $(1,2)$
So slope will be $-2$
Equation of normal will be $(y-2)=-2(x-1)$