Question

The x co-ordinate of a point on the curve 9y^2 = x^3 , the normal at which cuts off equal intercept on the axes is ?

Answer 1

9y^2=x^3
d(9y^2)/dx =d(x^3)/dx
18y dy/dx =3x^2
dy/dx =x^2/(6y)

Answer 2

The gradient of tangent is x^2/(6y)
So, gradient of normal is -6y/(x^2)

Answer 3

NOW, let x-intercept of the normal be a then y-intercept is also a.

Answer 4

Gradient of normal = (a-0)/(0-a)
-6y/(x^2) =-1

Answer 5

Now,
and solve for x