The Cartesian form of the curve given by x = | vedclass

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(B) Given equations are:$x = \frac{a}{2} (t + \frac{1}{t}) \dots (i)$$y = \frac{a}{2} (t - \frac{1}{t}) \dots (ii)$Squaring both equations and subtrac

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$x^2 - y^2 = a^2$

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(B) Given equations are:$x = \frac{a}{2} (t + \frac{1}{t}) \dots (i)$$y = \frac{a}{2} (t - \frac{1}{t}) \dots (ii)$Squaring both equations and subtracting $(ii)$ from $(i)$:$x^2 - y^2 = \frac{a^2}{4} (t + \frac{1}{t})^2 - \frac{a^2}{4} (t - \frac{1}{t})^2$$= \frac{a^2}{4} [(t^2 + \frac{1}{t^2} + 2) - (t^2 + \frac{1}{t^2} - 2)]$$= \frac{a^2}{4} [t^2 + \frac{1}{t^2} + 2 - t^2 - \frac{1}{t^2} + 2]$$= \frac{a^2}{4} (4) = a^2$Therefore,the Cartesian form is $x^2 - y^2 = a^2$.

The Cartesian form of the curve given by $x = \frac{a}{2} (t + \frac{1}{t})$ and $y = \frac{a}{2} (t - \frac{1}{t})$,where $t$ is a parameter,is:

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