Browse Questions

# Find the equation of the normal at the point $(am^2,am^3)$ for the curve $ay^2 = x^3.$

Toolbox:
• Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
• Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1) Step 1: Equation of the curve is ay^2=x^3 Differentiating w.r.t x we get, 2ay\large\frac{dy}{dx}$$=3x^2$
$\Rightarrow \large\frac{dy}{dx}=\frac{3x^2}{2ay}$
Step 2:
At the point $(am^2,am^3)$ the slope is given by
$\large\frac{dy}{dx}_{(am^2,am^3)}=\large\frac{3(am^2)^2}{2a(am^3)}$
$\qquad\qquad\;\;\;\;=\large\frac{3a^2m^4}{2a^2m^3}$
$\qquad\qquad\;\;\;\;=\large\frac{3m}{2}$
Therefore slope of the tangent at $(am^2,am^3)=\large\frac{3m}{2}$
Slope of the normal at $(am^2,am^3)=\large\frac{-2}{3m}$
Step 3: