Question

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 :

• A. $-2t^3$
• B. $-t^3$
• C. $0$
• D. $2t^3$

Answer 1

The correct option is A $-2t^3$

Equation of tangent at $P(t,t^3)$
$(y-t^3)=3t^2(x-t)$
Now solve the above equation with
$y=x^3$
We have
$x^3-t^3=3t^2(x-t)$
$\Rightarrow (x-t)(x^2+xt+t^2)=3t^2(x-t)$
$\Rightarrow x^2+xt-2t^2=0$
$\Rightarrow (x-t)(x+2t)=0$
$\Rightarrow x=-2t\Rightarrow Q(-2t,-8t^3)$
Ordinate of required point
$=\frac{2t^3+(-8t^3)}{3}=-2t^3$