A point on the parabola y^2 = 18x at which th | vedclass

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(A) Given the equation of the parabola: $y^2 = 18x$. Differentiating both sides with respect to $t$: $2y \left( \frac{dy}{dt} \right) = 18 \left( \fra

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$\left( \frac{9}{8}, \frac{9}{2} \right)$

Vedclass · Answer

(A) Given the equation of the parabola: $y^2 = 18x$. Differentiating both sides with respect to $t$: $2y \left( \frac{dy}{dt} ight) = 18 \left( \frac{dx}{dt} ight)$. We are given that the ordinate $(y)$ increases at twice the rate of the abscissa $(x)$,so $\frac{dy}{dt} = 2 \frac{dx}{dt}$. Substituting this into the differentiated equation: $2y \left( 2 \frac{dx}{dt} ight) = 18 \left( \frac{dx}{dt} ight)$. Assuming $\frac{dx}{dt} 
eq 0$,we get: $4y = 18 \implies y = \frac{18}{4} = \frac{9}{2}$. Now,substitute $y = \frac{9}{2}$ into the original parabola equation to find $x$: $\left( \frac{9}{2} ight)^2 = 18x$ $\frac{81}{4} = 18x$ $x = \frac{81}{4 	imes 18} = \frac{81}{72} = \frac{9}{8}$. Thus,the required point is $\left( \frac{9}{8}, \frac{9}{2} ight)$.

$A$ point on the parabola $y^2 = 18x$ at which the ordinate increases at twice the rate of the abscissa is

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