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

(A) $\frac{y}{x}$

(B) $-\frac{y}{x}$

(C) $\frac{x}{y}$

(D) $-\frac{x}{y}$

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Best answer

SOLUTION —

$\begin{array}{l}x^4+y^4=t^2+\frac{1}{t^2}=\left(t+\frac{1}{t}\right)^2-2 \\=\left(x^2+y^2\right)^2-2 \\\Rightarrow \quad x^4+y^4=x^4+y^4+2 x^2 y^2-2 \\\Rightarrow \quad x^2 y^2=1 \\\Rightarrow \quad 2 x \cdot y^2+2 x^2 y \frac{d y}{d x}=0 \\\Rightarrow \quad \frac{d y}{d x}=-\frac{y}{x} \\\end{array}$

So, The correct option will be (B).

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