Application of Derivatives Part 2

Given, curve ⇒y=x3
P(t,t3)
Equation of tangent at P(t,t3)
(y−t3)=3t2(x−t)
From Eqs. (i) and (ii),
x3−t3‌‌=3t2(x−t)
⇒‌(x−t)(x2+t2+xt)=3t2(x−t)
⇒‌x2+xt−2t2=0
⇒‌(x−t)(x+2t)=0
⇒‌x=t‌ or ‌x=−2t
This is not possible. Now, the coordinate of Q=(x,y)=(−2t,(−2t)3)
∴‌‌Q=(−2t,−8t3)
∴ Ordinate of the point dividing PQ in the ratio 1:2 is
‌

2t3+(−8t3)

1+2

=−2t3