If x = a(t - frac{1}{t}) and y = a(t + frac{1 | vedclass

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(C) Given equations are $x = a(t - \frac{1}{t})$ $(i)$ and $y = a(t + \frac{1}{t})$ $(ii)$.Squaring both equations,we get $x^2 = a^2(t^2 - 2 + \frac{1

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$\frac{x}{y}$

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(C) Given equations are $x = a(t - \frac{1}{t})$ $(i)$ and $y = a(t + \frac{1}{t})$ $(ii)$.Squaring both equations,we get $x^2 = a^2(t^2 - 2 + \frac{1}{t^2})$ and $y^2 = a^2(t^2 + 2 + \frac{1}{t^2})$.Subtracting $(i)^2$ from $(ii)^2$,we get $y^2 - x^2 = a^2(t^2 + 2 + \frac{1}{t^2} - (t^2 - 2 + \frac{1}{t^2})) = a^2(4) = 4a^2$.Differentiating $y^2 - x^2 = 4a^2$ with respect to $x$,we get $2y \frac{dy}{dx} - 2x = 0$.Therefore,$2y \frac{dy}{dx} = 2x$,which simplifies to $\frac{dy}{dx} = \frac{x}{y}$.

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

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