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(D) Given the parametric equations $x = \frac{t - 1}{t + 1}$ and $y = \frac{t + 1}{t - 1}$.At $t = 2$,$x = \frac{2 - 1}{2 + 1} = \frac{1}{3}$ and $y =

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$9x + y - 6 = 0$

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(D) Given the parametric equations $x = \frac{t - 1}{t + 1}$ and $y = \frac{t + 1}{t - 1}$.At $t = 2$,$x = \frac{2 - 1}{2 + 1} = \frac{1}{3}$ and $y = \frac{2 + 1}{2 - 1} = 3$.Observe that $xy = \left(\frac{t - 1}{t + 1}ight) 	imes \left(\frac{t + 1}{t - 1}ight) = 1$.Thus,$y = \frac{1}{x}$,which implies $\frac{dy}{dx} = -\frac{1}{x^2}$.At $x = \frac{1}{3}$,the slope $m = \frac{dy}{dx} = -\frac{1}{(1/3)^2} = -9$.The equation of the tangent at $(\frac{1}{3}, 3)$ is given by $y - y_1 = m(x - x_1)$.$y - 3 = -9(x - \frac{1}{3})$.$y - 3 = -9x + 3$.$9x + y - 6 = 0$.

The equation of the tangent to the curve $x = \frac{t - 1}{t + 1}, y = \frac{t + 1}{t - 1}$ at $t = 2$ is:

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