If $x^2+y^2=t-\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$, then $\frac{d y}{d x}=$

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Aicharya · Accepted Answer

Given, $\begin{aligned} & x^4+y^4=t^2+\frac{1}{t^2} \quad \ldots (i) \ & x^2+y^2=t-\frac{1}{t} \end{aligned}$ On Squaring both sides, we get $\begin{aligned} x^4+y^4+2 x^2 y^2 & =t^2+\frac{1}{t^2}-2 \ t^2+\frac{1}{t^2}+2 x^2 y^2 & =t^2+\frac{1}{t^2}-2 \quad[\because \text {from Eq. (i)}] \ x^2 y^2 & =-1 \ y^2 & =\frac{-1}{x^2} \end{aligned}$ Differentiating w.r.t. $x$, we get $\begin{aligned} 2 y y^{\prime} & =\frac{2}{x^3} \ y^{\prime} & =\frac{1}{x^3 y} \end{aligned}$ $\therefore$ Hence, answer is (d).

If \(x^2+y^2=t-\frac{1}{t}\) and \(x^4+y^4=t^2+\frac{1}{t^2}\), then \(\frac{d y}{d x}=\)

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