Question

The tangent to the curve $y = x^3$ at the point $P(t, t^3)$ meets the curve again at Q, then the ordinate of the point which divides PQ internally in the ratio 1:2 is :

• A. 0
• B. $2t^3$
• C. $-t^3$
• D. $-2t^3$

Hint:

For the curve $y = x^3$, the tangent at $x = t$ always intersects the curve again at $x = -2t$.

Answer 1

The Correct Option is D

Step 1: Slope of tangent at $P$ is $m = 3t^2$.
Step 2: Tangent equation: $y - t^3 = 3t^2(x-t) \Rightarrow y = 3t^2x - 2t^3$.
Step 3: Intersection with $y=x^3$: $x^3 - 3t^2x + 2t^3 = 0 \Rightarrow (x-t)^2(x+2t) = 0$.
Step 4: Point $Q$ has $x = -2t$. $y = (-2t)^3 = -8t^3$.
Step 5: Point $R$ dividing $PQ$ ($P(t, t^3), Q(-2t, -8t^3)$) in 1:2 ratio.
Step 6: Ordinate $y_R = \frac{1(-8t^3) + 2(t^3)}{3} = \frac{-6t^3}{3} = -2t^3$.