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

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Application of Derivatives

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:

Answer: (C)

The tangent to

y = x^{3}

P (t, t^{3})

has slope

3 t^{2}

, so its equation is

y - t^{3} = 3 t^{2} (x - t)

. Setting

y = x^{3}

and dividing out the known root at

x = t

leaves

x^{2} + x t - 2 t^{2} = 0

, that is

(x - t) (x + 2 t) = 0

, so the tangent meets the curve again at

Q (- 2 t, - 8 t^{3})

. The point dividing

P Q

internally in the ratio

1 : 2

has ordinate

\frac{1 \cdot ( - 8 t ^{3} ) + 2 \cdot t ^{3}}{1 + 2} = \frac{- 6 t ^{3}}{3} = - 2 t^{3}

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Single correct option