Question

Find the equation of the normal to the curve $y^2={ax}^{3}at(a,a)$

Answer 1

Slop of normal = $-1/dy/dx$

As $y^2=a{x}^2$

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

$\Rightarrow \frac{dy}{dx}=\frac{3a{x}^2}{2y}$ at $a,a=\frac{3a^2}{2a}=\frac{3}{2}{a}^2$

$\Rightarrow -1/\left(\frac{dy}{dx}\right)=\frac{-2}{3a^2}$

$\Rightarrow$Equation of normal

$y-a=\frac{-2}{3a^2}(x-a)$

$3a^{2}y-3a^3=-2x+2a$

At$(a,a)a^2=a^4\Rightarrow m=\frac{-2}{3a^2}$

$\Rightarrow$Equation $3y-3=-2x+2$ or $3y+3=-2x-2$