A point on the parabola y^2 = frac{36}{5}x at | vedclass

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Question

(B) Given the parabola equation: $y^2 = \frac{36}{5}x$. Differentiating both sides with respect to time $t$: $2y \frac{dy}{dt} = \frac{36}{5} \frac{dx

Vedclass · Accepted Answer

$(1/5, 6/5)$

Vedclass · Answer

(B) Given the parabola equation: $y^2 = \frac{36}{5}x$. Differentiating both sides with respect to time $t$: $2y \frac{dy}{dt} = \frac{36}{5} \frac{dx}{dt}$. We are given that the ordinate $(y)$ increases at thrice the rate of the abscissa $(x)$, so $\frac{dy}{dt} = 3 \frac{dx}{dt}$. Substituting this into the differentiated equation: $2y(3 \frac{dx}{dt}) = \frac{36}{5} \frac{dx}{dt}$. Assuming $\frac{dx}{dt} 
eq 0$, we get $6y = \frac{36}{5}$, which simplifies to $y = \frac{6}{5}$. Now, substitute $y = \frac{6}{5}$ into the original parabola equation: $(\frac{6}{5})^2 = \frac{36}{5}x$. $\frac{36}{25} = \frac{36}{5}x$. Solving for $x$: $x = \frac{36}{25} 	imes \frac{5}{36} = \frac{1}{5}$. Thus, the required point is $(1/5, 6/5)$.

$A$ point on the parabola $y^2 = \frac{36}{5}x$ at which the ordinate increases at thrice the rate of the abscissa is:

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