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The point on the curve y2=16xy^2 = 16x for which the y-coordinate is changing 2 times as fast as the x-coordinate is :

Solution

✅ Correct Option: 3

Differentiating: 2ydydt=16dxdt2y\frac{dy}{dt} = 16\frac{dx}{dt}, so dydt=8ydxdt\frac{dy}{dt} = \frac{8}{y}\frac{dx}{dt}.

Given dydt=2dxdt\frac{dy}{dt} = 2\frac{dx}{dt}: 8y=2⇒y=4\frac{8}{y} = 2 \Rightarrow y = 4.

x=y216=1x = \frac{y^2}{16} = 1. Point (1,4)(1, 4).

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