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If x2+y2=t+1tx^2+y^2 = t+\dfrac{1}{t} and x4+y4=t2+1t2x^4+y^4=t^2+\dfrac{1}{t^2}, then dydx\dfrac{dy}{dx} is equal to

Solution

✅ Correct Option: 2

Squaring the first relation, (x2+y2)2=t2+2+1t2=(x4+y4)+2(x^2+y^2)^2=t^2+2+\dfrac{1}{t^2}=(x^4+y^4)+2, so 2x2y2=22x^2y^2=2, i.e. x2y2=1x^2y^2=1. Differentiating, 2xy2+2x2ydydx=02xy^2+2x^2y\dfrac{dy}{dx}=0, giving dydx=−yx\dfrac{dy}{dx}=-\dfrac{y}{x}.

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