JEE PYQ: Application Of Derivatives Question 11
Question 11 - 2021 (24 Feb Shift 1)
If 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:
(1) $-2t^3$
(2) $-t^3$
(3) $0$
(4) $2t^3$
Show Answer
Answer: (1)
Solution
Tangent at $P(t, t^3)$: $y - t^3 = 3t^2(x - t)$. Substituting $y = x^3$: $x^3 - t^3 = 3t^2(x - t)$ gives $(x - t)(x^2 + xt + t^2 - 3t^2) = 0$, so $x = -2t$. Thus $Q(-2t, -8t^3)$. Ordinate dividing $PQ$ in ratio $1:2 = \frac{2t^3 + (-8t^3)}{3} = -2t^3$.
BETA
