Question

Find the equation of the normal at the point (am2, am3 ) for the curve ay2 = x3.

Answer 1

The equation of the given curve is ay2 = x3 .

On differentiating with respect to x, we have:

2ay$\frac{dy}{dx}$=3x2

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

The slope of a tangent to the curve at (x0, y0) is $\frac{dy}{dx}$](x0,y0).

The slope of the tangent to the given curve at (am2 , am3 ) is

$\frac{dy}{dx}$(am2,am3)=$\frac{3(m^2)^2}{2a(am^3)}$=$\frac{3a^2m^4}{2a^2m^3}$=$\frac{3m}{2}$.

Slope of normal at (am2 , am3 ) is given by,

y-am3=$-\frac{2}{3}$m(x-am2)

3my-3am4=-2x+2am2

2x+3my-am2(2+3m2)=0