If $x^2+y^2=\mathrm{t}+\frac{1}{\mathrm{t}}$ and $x^4+y^4=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ is equal to

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Aicharya · Accepted Answer

$x^2+y^2=\mathrm{t}+\frac{1}{\mathrm{t}}$ Squaring on both sides, we get $x^4+y^4+2 x^2 y^2=t^2+\frac{1}{t^2}+2$ $\begin{aligned} 	herefore \quad \mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 x^2 y^2=\mathrm{t}^2 & +\frac{1}{\mathrm{t}^2}+2 \ & \ldots\left[\because x^4+y^4=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}, 	ext { given }ight]\end{aligned}$ $\begin{aligned} & 2 x^2 y^2=2 \ & x^2 y^2=1 \end{aligned}$ differentiating w.r.t. $x$, we get $\begin{aligned} & x^2 2 y \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x y^2=0 \ & x^2 2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=-2 x y^2 \ & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-2 x y^2}{2 x^2 y}=\frac{-y}{x} \end{aligned}$

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

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