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

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Aicharya · Accepted Answer

We have, $x^{2}+y^{2}=t+\frac{1}{t} \cdots$ (1) $\quad$ and $\quad x^{4}+y^{4}=t^{2}+\frac{1}{t^{2}}$...(2) Squaring (1), we get $\begin{array}{l} x^{4}+y^{4}+2 x^{2} y^{2}=t^{2}+\frac{1}{t^{2}}+2 \Rightarrow x^{4}+y^{4}+2 x^{2} y^{2}=x^{4}+y^{4}+2 	ext {..(from (2)) } \ 	herefore x^{2} y^{2}=1 \end{array}$ Differentiating w.r.t. $x$, we get $\begin{aligned} & x^{2}\left(2 y \frac{d y}{d x}ight)+y^{2}(2 x)=0 \ 	herefore & \frac{d y}{d x}=\frac{-2 x y^{2}}{2 x^{2} y}=\frac{-y}{x} \end{aligned}$

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

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