question_answer

Find the point on the parabola \[{{y}^{2}}=18x\] at which the ordinate increases at twice the rate of the abscissa.

Answer:

Given,   \[{{y}^{2}}=18x\]                               ?(i) On differentiating both sides w.r.t. t, we get             \[2y\frac{dy}{dt}=18\frac{dx}{dt}\]                               ?(ii) According to the given condition,             \[\frac{dy}{dt}=2\frac{dx}{dt}\] \[\therefore \] from Eq (ii), we get \[2y\,\,\left[ 2\frac{dx}{dt} \right]=18\frac{dx}{dt}\,\,\Rightarrow \,\,4y=18\Rightarrow y=\frac{9}{2}\] On putting \[y=\frac{9}{2}\] in Eq. (i), we get             \[{{\left( \frac{9}{2} \right)}^{2}}=18x\] \[\Rightarrow \] \[x=\frac{81}{4\times 18}=\frac{9}{8}\] Hence, the required point on the curve is \[\left( \frac{9}{8},\,\,\frac{9}{2} \right).\]