If {x^2} + {y^2} = t - frac{1}{t} and {x^4} + | 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 $(1)$,we get:$(x^2 + y^2

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$1$

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(A) Given equations are:$x^2 + y^2 = t - \frac{1}{t}$  --- $(1)$$x^4 + y^4 = t^2 + \frac{1}{t^2}$  --- $(2)$Squaring equation $(1)$,we get:$(x^2 + y^2)^2 = (t - \frac{1}{t})^2$$x^4 + y^4 + 2x^2y^2 = t^2 + \frac{1}{t^2} - 2$Substitute the value from equation $(2)$ into this expression:$(t^2 + \frac{1}{t^2}) + 2x^2y^2 = t^2 + \frac{1}{t^2} - 2$$2x^2y^2 = -2$$x^2y^2 = -1$Since $x^2y^2 = -1$,we have $y^2 = -\frac{1}{x^2}$.Differentiating both sides with respect to $x$:$2y \frac{dy}{dx} = \frac{d}{dx}(-x^{-2})$$2y \frac{dy}{dx} = -(-2)x^{-3}$$2y \frac{dy}{dx} = \frac{2}{x^3}$$y \frac{dy}{dx} = \frac{1}{x^3}$Multiplying both sides by $x^3$,we get:$x^3y \frac{dy}{dx} = 1$.

If ${x^2} + {y^2} = t - \frac{1}{t}$ and ${x^4} + {y^4} = {t^2} + \frac{1}{t^2}$,then ${x^3}y\frac{dy}{dx} = $

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