Question

Tangent at a point P₁ (other than (0, 0) on the curve y = x³ meets the curve at P₂ and tangent at P₂ meets the curve at P₃ and so on. Show that the abcissae of P₁, P₂, ..., P_n form a G.P. Also, find the ratio of (Ar(Δ P₁P₂P₃)/Ar(ΔP₂P₃P₄))

Answer 1

Let P₁(x₁, y₁) be a point on a curve y = x³

Now, dy/dx = 3x²

Slope of the tangent at P₁ = m₁ = 3x₁²

Equation of the tangent at P₁ is

y – y₁ = 3x₁²(x – x1)

⇒ x³ – x₁³ = 3x₁² (x – x₁)

⇒ x₃ – 3x₁² x – 2x₁³ = 0

⇒ (x – x₁)(x² + x x₁ – 2x₁²) = 0

⇒ (x – x₁)(x – x₁)(x + 2x₁) = 0

⇒ x = x₁, – 2x₁

⇒ x = – 2x₁

Thus, x₂ = –2x₁, y₂ = x₂³ = – 8x₁³

Therefore, P₂ = (x₂, y₂) = (–2x₁, –8x₁³)

Now, we find P₃.

Equation of tangent at P₂ is

⇒ y – y₂ = 3x₂²(x – x²)

⇒ x³ – x₂³ = 3x₂² (x – x²)

⇒ (x – x₂)(x – x₂)(x + 2x₂) = 0

⇒ x = – 2x₂ 

Thus, x₃ = – 2x₂ = –2. – 2x₁ = 4x₁

and y₃ = x₂³ = (4x₁)³ = 64x₁³

Therefore, P₃ = (x₃, y₃) = (4x₁, 64x₁³)

The abscissae of P₁, P₂, P₃, ... are x₁, – 2x₁, 4x₁, – 8x₁

Which are in G.P with a common ratio -2.

Now, ar(ΔP₁P₂P₃)