Which of the following equations in parametri | vedclass

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(D) For option $A$: Given $x = \frac{a}{2}(t + \frac{1}{t})$ and $y = \frac{b}{2}(t - \frac{1}{t})$.Squaring both,we get $\frac{4x^2}{a^2} = (t + \fra

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All of the above

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(D) For option $A$: Given $x = \frac{a}{2}(t + \frac{1}{t})$ and $y = \frac{b}{2}(t - \frac{1}{t})$.Squaring both,we get $\frac{4x^2}{a^2} = (t + \frac{1}{t})^2 = t^2 + \frac{1}{t^2} + 2$ and $\frac{4y^2}{b^2} = (t - \frac{1}{t})^2 = t^2 + \frac{1}{t^2} - 2$.Subtracting the two: $\frac{4x^2}{a^2} - \frac{4y^2}{b^2} = 4$,which simplifies to $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,representing a hyperbola.For option $B$: Given $x^2 - 6 = 2 \cos t$ and $y^2 + 2 = 4 \cos^2 \frac{t}{2}$.Using $2 \cos^2 \frac{t}{2} = 1 + \cos t$,we have $y^2 + 2 = 2(1 + \cos t) = 2 + 2 \cos t$,so $y^2 = 2 \cos t$.Substituting $2 \cos t = x^2 - 6$,we get $y^2 = x^2 - 6$,or $x^2 - y^2 = 6$,which represents a rectangular hyperbola.For option $C$: Given $x = e^t + e^{-t}$ and $y = e^t - e^{-t}$.Then $x^2 - y^2 = (e^t + e^{-t})^2 - (e^t - e^{-t})^2 = (e^{2t} + e^{-2t} + 2) - (e^{2t} + e^{-2t} - 2) = 4$.This is $x^2 - y^2 = 4$,which also represents a rectangular hyperbola.Since all options represent hyperbolas,the correct answer is $D$.

Which of the following equations in parametric form can represent a hyperbola,where $t$ is a parameter?

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