Question

If the equation of the normal of the curve x^2/3 + y^2/3 = a^2/3 makes an angle f with the positive direction of x-axis, prove that the equation of the normal x sinϕ – y cosϕ + acos2ϕ = 0.

Answer 1

Given curve is x^2/3 + y^2/3 = a^2/3 ...(i)

Differentiating w.r.t x, we get,

2/3x^–1/3 + 2/3 y^–1/3 dy/dx = 0

⇒ dy/dx = – y^1/3/x^1/3

Let P(α, β) be any point on the curve (i)

Slope of the normal at P is m = (α/β)^1/3 

It is given that, the slope of the normal = tanϕ

Hence, the equation of the normal is