If x=aleft(t-frac{1}{t}right) and y=bleft(t+f | vedclass

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(D) Given $x=a\left(t-\frac{1}{t}\right)$ and $y=b\left(t+\frac{1}{t}\right)$.Differentiating with respect to $t$:$\frac{dx}{dt}=a\left(1+\frac{1}{t^2

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$\frac{b^2 x}{a^2 y}$

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(D) Given $x=a\left(t-\frac{1}{t}ight)$ and $y=b\left(t+\frac{1}{t}ight)$.Differentiating with respect to $t$:$\frac{dx}{dt}=a\left(1+\frac{1}{t^2}ight) = a\left(\frac{t^2+1}{t^2}ight)$$\frac{dy}{dt}=b\left(1-\frac{1}{t^2}ight) = b\left(\frac{t^2-1}{t^2}ight)$Using the chain rule,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{b\left(\frac{t^2-1}{t^2}ight)}{a\left(\frac{t^2+1}{t^2}ight)} = \frac{b(t^2-1)}{a(t^2+1)}$.From the given equations:$\frac{x}{a} = t-\frac{1}{t} = \frac{t^2-1}{t}$$\frac{y}{b} = t+\frac{1}{t} = \frac{t^2+1}{t}$Dividing these two expressions:$\frac{x/a}{y/b} = \frac{(t^2-1)/t}{(t^2+1)/t} = \frac{t^2-1}{t^2+1}$Substituting this into the expression for $\frac{dy}{dx}$:$\frac{dy}{dx} = \frac{b}{a} 	imes \left(\frac{x/a}{y/b}ight) = \frac{b}{a} 	imes \frac{xb}{ya} = \frac{b^2 x}{a^2 y}$.

If $x=a\left(t-\frac{1}{t}\right)$ and $y=b\left(t+\frac{1}{t}\right)$,then $\frac{dy}{dx}=$

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