Find the equation of normal to the curve $ay^2 = x^3$ at the point whose x-coordinate is $am^2$.

$2x+3my−3am^4-2am^2=0$

The correct answer is Option (2) → $2x+3my−3am^4-2am^2=0$

The given curve is $ay^2 = x^3$   ...(i)

Putting $x = am^2$ in (i), we get

$ay^2 = (am^2)^3 ⇒ ay^2 = a^3m^6⇒ y^2 = a^2m^6⇒ y = ±am^3$.

Hence, the feet of normals to the curve (i) are $(am^2, am^3)$ and $(am^2, - am^3)$.

Differentiating (i) w.r.t. x, we get

$2ay\frac{dy}{dx}=3x^2⇒\frac{dy}{dx}=\frac{3x^2}{2ay}$

Normal at $(am^2, am^3)$

Slope of tangent at $(am^2, am^3) = \left(\frac{dy}{dx}\right)_{(am^2, am^3)}=\frac{3(am^2)^2}{2a.am^3}=\frac{3m}{2}$

⇒ slope of normal at $(am^2, am^3)=-\frac{3m}{2}$

The equation of normal at $(am^2, am^3)$ is

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

$⇒2x+3my−3am^4-2am^2=0$

Normal at $(am^2, -am^3)$

Slope of tangent at $(am^2, -am^3) = \left(\frac{dy}{dx}\right)_{(am^2, -am^3)}=\frac{3(am^2)^2}{-2a.am^3}=-\frac{3m}{2}$

⇒ slope of normal at $(am^2, -am^3)=\frac{3m}{2}$

The equation of normal at $(am^2, -am^3)$ is

$y+am^3=\frac{3m}{2}(x-am^2)$

$⇒2x-3my−3am^4-2am^2=0$

Hence, the required equations of normals are $2x + 3my - 3am^4 - 2am^2 = 0$ and $2x-3my - 3am^4-2am^2 = 0$.