The tangent to the curve $y=x^3$ at the point $P\left(t, t^3\right)$ cuts the curve again at point Q. Then, the coordinates of Q are
Answer & explanation
Correct answer: option 4
Let the coordinates of Q be $\left(t_1, t_1{ }^3\right)$. Then,
Slope of PQ = $\frac{t_1{ }^3-t^3}{t_1-t}=t_1{ }^2+t t_1+t^2$
Also,
Slope of PQ = $\left(\frac{d y}{d x}\right)_P=3 t^2$
∴ $3 t^2=t_1{ }^2+t t_1+t^2$
$\Rightarrow t_1{ }^2+t t_1-2 t^2=0$
$\Rightarrow \left(t_1+2 t\right)\left(t_1-t\right)=0 \Rightarrow t_1=-2 t$ [∵ t1 = t given point P]
Thus, the coordinates of Q are $\left(-2 t,-8 t^3\right)$.