Question

Find the equation of the normal at the point ($a{m}^2,a{m}^3$ ) for the curve $a{y}^2=x^3$ .

Answer 1

Given equation of curve is

$a{y}^2=x^3$
Differentiating w.r.t. $x$, we get
$2ay\frac{dy}{dx}=3x^2$
$\Rightarrow \frac{dy}{dx}=\frac{3x^2}{2ay}$

Slope of the tangent to the curve at $(a{m}^2,a{m}^3)$ is
${\left(\begin{array}{c}\frac{dy}{dx}\end{array}\right)}_{(a{m}^2,a{m}^3)}=\frac{3(a{m}^2{)}^2}{2a\left(a{m}^3\right)}=\frac{3a^2{m}^4}{2a^2{m}^3}=\frac{3m}{2}$

Slope of normal at $(a{m}^2,a{m}^3)$
$=\frac{-1}{\text{slope of the tangent at}(a{m}^2,a{m}^3)}=\frac{-2}{3m}$

Equation of the normal at $(a{m}^2,a{m}^3)$ is
$y-a{m}^3=\frac{-2}{3m}(x-a{m}^2)$
$\Rightarrow 3my-3a{m}^4=-2x+2a{m}^2$
$\Rightarrow 2x+3my-a{m}^2(2+3m^2)=0$