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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:
Option: 1 -2t^{3}
Option: 2 -t^{3}
Option: 3 0  
Option: 4 2t^{3}

Answers (1)

best_answer

 

equation of tangent at \mathrm{P}\left(\mathrm{t},\mathrm{t}^{3}\right)

\\\\ \left(y-t^{3}\right)=3 t^{2}(x-t)\qquad\ldots(1)

now solve the above equation with

y=x^{3}\qquad\ldots(2)

\\\text{By }(1) \& (2)\\ \\x^{3}-t^{3}=3 t^{2}(x-t)\\ \\\mathrm{x}^{2}+\mathrm{xt}+\mathrm{t}^{2}=3 \mathrm{t}^{2}\\ \\\mathrm{x}^{2}+\mathrm{xt}-2 \mathrm{t}^{2}=0\\ \\(x-t)(x+2 t)=0\\ \\\Rightarrow x=-2 t \Rightarrow Q\left(-2 t,-8 t^{3}\right)\\ \\\text{Ordinate of required point} \\\\ =\frac{2 t^{3}+\left(-8 t^{3}\right)}{3}=-2 t^{3}

Posted by

Suraj Bhandari

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