If x^{2}+y^{2}=t+frac{1}{t} and x^{4}+y^{4}=t | vedclass

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(A) Given equations are:$x^{2}+y^{2}=t+\frac{1}{t}$ ... $(1)$$x^{4}+y^{4}=t^{2}+\frac{1}{t^{2}}$ ... $(2)$Squaring equation $(1)$ on both sides,we get

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$-\frac{y}{x}$

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(A) Given equations are:$x^{2}+y^{2}=t+\frac{1}{t}$ ... $(1)$$x^{4}+y^{4}=t^{2}+\frac{1}{t^{2}}$ ... $(2)$Squaring equation $(1)$ on both sides,we get:$(x^{2}+y^{2})^{2} = (t+\frac{1}{t})^{2}$$x^{4}+y^{4}+2x^{2}y^{2} = t^{2}+\frac{1}{t^{2}}+2$Substituting the value from equation $(2)$ into this result:$(t^{2}+\frac{1}{t^{2}}) + 2x^{2}y^{2} = t^{2}+\frac{1}{t^{2}}+2$$2x^{2}y^{2} = 2$$x^{2}y^{2} = 1$Differentiating both sides with respect to $x$ using the product rule:$x^{2}(2y \frac{dy}{dx}) + y^{2}(2x) = 0$$2x^{2}y \frac{dy}{dx} = -2xy^{2}$$\frac{dy}{dx} = \frac{-2xy^{2}}{2x^{2}y} = -\frac{y}{x}$

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

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