Find a point on the curve y = x 3 – 3x where the tangent is parallel to the chord joining (1, – 2) and (2, 2).

Question

Philoid · Accepted Answer

Given:

The curve y = x³ – 3x

First, we will find the Slope of the tangent

y = x³ – 3x

(x³) – (3x)

⇒ = 3x^{3 – 1} – )

⇒ = 3x² – 3 ...(1)

The equation of line passing through (x₀,y₀) and The Slope m is y – y₀ = m(x – x₀).

so The Slope, m =

The Slope of the chord joining (1, – 2) & (2,2)

⇒

⇒ = 4 ...(2)

From (1) & (2)

3x² – 3 = 4

⇒ 3x² = 7

⇒ x² =

⇒ x =

y = x³ – 3x

⇒ y = x(x² – 3)

⇒ y = (()² – 3)

⇒ y = (( – 3)

⇒ y = ()

Thus, the required point is x = & y = ()