If the tangent to the curve y = x3 at the point P(t, t3) meets the curve again at Q, then the
ordinate of the point which divides PQ internally in the ratio 1 : 2 is :
Solutions · 1
Official solution
Given P(t,t3)
Let Q=(t1,t13)
Slope of tangent at point p,
\[{{dy} \over {dx}} = 3{x^2}\]
\[ \Rightarrow {\left. {{{dy} \over {dx}}} \right|_{(t,{t^3})}} = 3{t^2}\]
This slope is same as slope of line PQ.
Slope of \[PQ = {{t_1^3 - {t_3}} \over {{t_1} - t}}\]
\[ \therefore \] \[3{t^2} = {{t_1^3 - {t_3}} \over {{t_1} - t}}\]
\[ \Rightarrow 3{t^2} = {{({t_1} - t)(t_1^2 + t{t_1} + {t^2})} \over {({t_1} - t)}}\]
\[ \Rightarrow 3{t^2} = t_1^2 + t{t_1} + {t^2}\]
\[ \Rightarrow {t_1} = - 2t\]
\[ \therefore \] \[Q = \left( { - 2t, - 8{t^3}} \right)\]
\[ \therefore \] \[h = {{2t - 2t} \over 3} = 0\]
\[k = {{2{t^3} - 8{t^3}} \over 3} = - 2{t^3}\]
\[ \therefore \] Point M=(0,−2t3)