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

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 $\mathrm{PQ}$ internally in the ratio $1: 2$ is :

1. (1) $-2 t^{3}$

2. (2) $-\mathrm{t}^{3}$

3. (3) 0

4. (4) $2 t^{3}$

Correct Option: 1,

Solution:

Equation of tangent at $\mathrm{P}\left(\mathrm{t}, \mathrm{t}^{3}\right)$

$\left(y-t^{3}\right)=3 t^{2}(x-t) \ldots$ (i)

Now solve the above equation with

$y=x^{3} \ldots$ (ii)

By (i) and (ii),

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

$x^{2}+x t+t^{2}=3 t^{2}$

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

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

$\Rightarrow x=-2 t \Rightarrow Q\left(-2 t,-8 t^{3}\right)$

Ordinate of required point $=\frac{2 t^{3}+\left(-8 t^{3}\right)}{3}=-2 t^{3}$