The point on the curve y^2 = 18x,at which the | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the curve equation is $y^2 = 18x$. Differentiating both sides with respect to $t$,we get: $2y \frac{dy}{dt} = 18 \frac{dx}{dt}$ $y \frac{dy}{dt} = 9 \frac{dx}{dt}$ According to the problem,the rate of change of $Y$-coordinate is twice the rate of change of $X$-coordinate,i.e.,$\frac{dy}{dt} = 2 \frac{dx}{dt}$. Substituting this into the differentiated equation: $y(2 \frac{dx}{dt}) = 9 \frac{dx}{dt}$ Since $\frac{dx}{dt} 
eq 0$,we can divide by $\frac{dx}{dt}$: $2y = 9 \implies y = \frac{9}{2}$. Now,substitute $y = \frac{9}{2}$ into the original curve equation $y^2 = 18x$: $\left(\frac{9}{2}ight)^2 = 18x$ $\frac{81}{4} = 18x$ $x = \frac{81}{4 	imes 18} = \frac{81}{72} = \frac{9}{8}$. Thus,the required point is $\left(\frac{9}{8}, \frac{9}{2}ight)$.

The point on the curve $y^2 = 18x$,at which the rate of change of $Y$-coordinate is twice the rate of change of $X$-coordinate is . . . . . . . $\left(\frac{dx}{dt} \neq 0\right)$

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