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

The equation of the given curve is $a{y}^2=x^3$ ....(i)

On differentiating both sides of Eq. (i) w.r.t x, we get

$a\left(2y\right)\frac{dy}{dx}=3x^2\Rightarrow \frac{dy}{dx}=\frac{3x^2}{2ay}$

The slope of the tangent to the given curve at $(a{m}^2,a{m}^3)$ is

${\left(\frac{dy}{dx}\right)}_{(a{m}^2,a{m}^3)}=\frac{3(a{m}^2{)}^2}{2a\left(a{m}^3\right)}=\frac{3}{2}m$

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

Hence, equation of normal at $(a{m}^2,a{m}^3)$ is given by

$y-a{m}^3=-\frac{2}{3m}(x-a{m}^2)\Rightarrow 3my-3a{m}^4=-2x+2a{m}^2$

$\Rightarrow 2x+3my-3a{m}^4-2a{m}^2=0$