Find the point on the parabola y^2 = 18x wher | vedclass

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(D) Given the equation of the parabola is $y^2 = 18x$. Differentiating both sides with respect to $t$,we get $2y \frac{dy}{dt} = 18 \frac{dx}{dt}$,whi

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

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(D) Given the equation of the parabola is $y^2 = 18x$. Differentiating both sides with respect to $t$,we get $2y \frac{dy}{dt} = 18 \frac{dx}{dt}$,which simplifies to $y \frac{dy}{dt} = 9 \frac{dx}{dt}$ (Equation $1$). It is given that the rate of change of the $y$-coordinate is twice the rate of change of the $x$-coordinate,i.e.,$\frac{dy}{dt} = 2 \frac{dx}{dt}$. Substituting this into Equation $1$,we get $y(2 \frac{dx}{dt}) = 9 \frac{dx}{dt}$. Assuming $\frac{dx}{dt} 
eq 0$,we have $2y = 9$,which implies $y = \frac{9}{2}$. Substituting $y = \frac{9}{2}$ into the original equation $y^2 = 18x$,we get $\left( \frac{9}{2} ight)^2 = 18x$. This gives $\frac{81}{4} = 18x$,so $x = \frac{81}{4 	imes 18} = \frac{9}{8}$. Thus,the required point is $\left( \frac{9}{8}, \frac{9}{2} ight)$.

Find the point on the parabola $y^2 = 18x$ where the rate of change of the $y$-coordinate is twice the rate of change of the $x$-coordinate.

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