Question

Find the equation of the normal to curve $x^3=4y$ which passes through the point $(1,2)$

Answer 1

$x^3=4y\Rightarrow 3x^2=4\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{3}{4}{x}^2$

$\therefore$ Slope of normal $=-\frac{4}{3x^2}$

$\therefore$ Equation of normal passing through $(1,2)$

$\Rightarrow (y-2)=-\frac{4}{3}\left(\frac{1}{x^2}\right)(x-1)$

at $x=1$, Slope of normal $=-\frac{4}{3}\left(\frac{1}{1^2}\right)=-\frac{4}{3}$

$\Rightarrow (y-2)=-\frac{4}{3}(x-1)\Rightarrow 3y-6=-4x+4\Rightarrow 3y+4x-10=0$