The tangent to the curve `y=x^(3)` at the point `P(t, t^(3))` cuts the curve again at point Q. Then, the coordinates of Q are

Options

A. (0, 0)
B. `(2t, 4t^(3))`
C. `(2t, 8t^(3))`
D. `(-2t, -8t^(3))`

Answer

D

Explanation

Let the coordinates of Q be `(t_(1),t_(1)^(3)).` Then,
Slope of `PQ=(t_(1)^(3)-t^(3))/(t_(1)-t)=t_(1)^(2)+t t_(1)+t^(2)`
Also,
Slope of `PQ=((dy)/(dx))_(p)=3t^(2)`
`therefore 3t^(2)=t_(1)^(2)+t t_(1)+t^(2)`
`rArr t_(1)^(2)+t t_(1)-2t^(2)=0`
`rArr (t_(1)+2t)(t_(1)-t)=0rArr t_(1)= -2t " "[because t_(1)=t" given point P"]`
Thus, the coordinates of Q are `(-2t, -8t^(3)).`