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Let x=at2,y=2atx=at^2, y=2at. If d2ydx2+dydx=0\frac{d^2y}{dx^2}+\frac{dy}{dx}=0 at t=3t=3, then aa is equal to:

Solution

✅ Correct Option: 2

dydx=2a2at=1t\frac{dy}{dx}=\frac{2a}{2at}=\frac1t and d2ydx2=−1/t22at=−12at3\frac{d^2y}{dx^2}=\frac{-1/t^2}{2at}=-\frac{1}{2at^3}. At t=3t=3: −154a+13=0-\frac{1}{54a}+\frac13=0, so 54a=354a=3 and a=118a=\frac{1}{18}.

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