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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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Verified Answer
The correct answer is:
$\frac{-y}{x}$
$x^2+y^2=\mathrm{t}+\frac{1}{\mathrm{t}}$
Squaring on both sides, we get
$x^4+y^4+2 x^2 y^2=t^2+\frac{1}{t^2}+2$
$\begin{aligned} \therefore \quad \mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 x^2 y^2=\mathrm{t}^2 & +\frac{1}{\mathrm{t}^2}+2 \\ & \ldots\left[\because x^4+y^4=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}, \text { given }\right]\end{aligned}$
$\begin{aligned}
& 2 x^2 y^2=2 \\
& x^2 y^2=1
\end{aligned}$
differentiating w.r.t. $x$, we get
$\begin{aligned}
& x^2 2 y \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x y^2=0 \\
& x^2 2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=-2 x y^2 \\
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-2 x y^2}{2 x^2 y}=\frac{-y}{x}
\end{aligned}$
Squaring on both sides, we get
$x^4+y^4+2 x^2 y^2=t^2+\frac{1}{t^2}+2$
$\begin{aligned} \therefore \quad \mathrm{t}^2+\frac{1}{\mathrm{t}^2}+2 x^2 y^2=\mathrm{t}^2 & +\frac{1}{\mathrm{t}^2}+2 \\ & \ldots\left[\because x^4+y^4=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}, \text { given }\right]\end{aligned}$
$\begin{aligned}
& 2 x^2 y^2=2 \\
& x^2 y^2=1
\end{aligned}$
differentiating w.r.t. $x$, we get
$\begin{aligned}
& x^2 2 y \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x y^2=0 \\
& x^2 2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=-2 x y^2 \\
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-2 x y^2}{2 x^2 y}=\frac{-y}{x}
\end{aligned}$
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