JEE Main 2021 (Online) 24th February Morning ShiftQuestion 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 :
Differentiation and its Application
Solution
Concept. Substituting the tangent into $y=x^3$ gives a cubic with a double root at $x=t$; the third root is $Q$. Formulas used. (1) Section formula: the point dividing $PQ$ in ratio $1:2$ has ordinate $\dfrac{1\cdot y_Q+2\cdot y_P}{3}$Step 1 - Tangent. $y-t^3=3t^2(x-t)$. Step 2 - Meet the curve. $$x^3-3t^2x+2t^3=0\ \Longrightarrow\ (x-t)^2(x+2t)=0.$$ So $Q=(-2t,-8t^3)$.Step 3 - Ordinate. By (1), $\dfrac{-8t^3+2t^3}{3}=-2t^3$. Answer: (D)