On the curve y = 4x³ – 2x⁵, find all points at which tangents pass through the origin.

Formula Used:

•Equation of the tangent at (x₁, y₁) where slope is m is given by y – y₁ = m(x – x₁)

•Equation of the normal at (x₁, y₁) where slope is m is given by

Let (x₁, y₁) be the required point on the given curve y = 4x³ – 2x⁵

⇒ y₁ = 4x₁³ – 2x₁⁵ … (1)

Differentiating wrt x we get,

Equation of the tangent is y – y₁ = (12x₁² – 10x₁⁴)(x – x₁)

This passes through the origin.

Hence, 0 – y₁ = (12x₁² – 10x₁⁴)(0 – x₁)

⇒ y₁ = 12x₁³ – 10x₁⁵ … (2)

Eq (1) - Eq (2),

0 = -8x₁³ + 8x₁⁵

⇒ x₁ = 0 or x₁ = ±1

When x₁ = 0, from eq (2), y₁ = 0

When x₁ = 1, from eq (2),

y₁ = 4 – 2 = 2

When x₁ = -1, from eq (2),

⇒ y₁ = -4 – (-2) = -2

Hence, the required points are (0, 0), (1, 2), (-1, -2)