Solve this following

Question:

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

1. $-2 \mathrm{t}^{3}$

2. 0

3. $-\mathrm{t}^{3}$

4. $2 \mathrm{t}^{3}$

Correct Option: 1

Solution:

Slope of tangent at $\left.P\left(t, t^{3}\right)=\frac{d y}{d x}\right]_{\left(t, t^{3}\right)}$

$=\left(3 x^{2}\right)_{x=t}=3 t^{2}$

So equation tangent at $\mathrm{P}\left(\mathrm{t}, \mathrm{t}^{3}\right)$ :

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

for point of intersection with $\mathrm{y}=\mathrm{x}^{3}$

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

$\Rightarrow \quad(x-t)\left(x^{2}+x t+t^{2}\right)=3 t^{2}(x-t)$

for $x \neq t$

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

$\Rightarrow x^{2}+x t-2 t^{2}=0 \Rightarrow(x-t)(x+2 t)=0$

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

ordinate of required point $: \frac{2 t^{3}-8 t^{3}}{2+1}=-2 t^{3}$