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(D) Given $x = \sec^2 t$ and $y = \cot t$.Since $\sec^2 t = 1 + 	an^2 t = 1 + \frac{1}{\cot^2 t} = 1 + \frac{1}{y^2}$,we have $x - 1 = \frac{1}{y^2}$

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$5$

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(D) Given $x = \sec^2 t$ and $y = \cot t$.Since $\sec^2 t = 1 + 	an^2 t = 1 + \frac{1}{\cot^2 t} = 1 + \frac{1}{y^2}$,we have $x - 1 = \frac{1}{y^2}$,which implies $y^2(x - 1) = 1$.At $t = \pi/4$,$x = \sec^2(\pi/4) = 2$ and $y = \cot(\pi/4) = 1$. So,$P = (2, 1)$.Differentiating $y^2(x - 1) = 1$ with respect to $x$,we get $2y \frac{dy}{dx}(x - 1) + y^2 = 0$.At $(2, 1)$,$2(1) \frac{dy}{dx}(2 - 1) + 1^2 = 0 \Rightarrow 2 \frac{dy}{dx} + 1 = 0 \Rightarrow \frac{dy}{dx} = -1/2$.The equation of the tangent at $P(2, 1)$ is $y - 1 = -\frac{1}{2}(x - 2) \Rightarrow 2y - 2 = -x + 2 \Rightarrow x + 2y = 4$.Substituting $y = \frac{4-x}{2}$ into the curve equation $y^2(x - 1) = 1$:$\left(\frac{4-x}{2}ight)^2(x - 1) = 1 \Rightarrow (4-x)^2(x - 1) = 4 \Rightarrow (16 - 8x + x^2)(x - 1) = 4$.$16x - 16 - 8x^2 + 8x + x^3 - x^2 = 4 \Rightarrow x^3 - 9x^2 + 24x - 20 = 0$.Since $P(2, 1)$ is on the curve,$(x - 2)^2$ must be a factor. Dividing by $(x - 2)^2 = x^2 - 4x + 4$,we get $(x - 2)^2(x - 5) = 0$.The roots are $x = 2$ (twice) and $x = 5$. Thus,the $x$-coordinate of $Q$ is $5$.

$A$ curve is represented by the equations $x = \sec^2 t$ and $y = \cot t$,where $t$ is a parameter. If the tangent at the point $P$ on the curve where $t = \pi/4$ meets the curve again at the point $Q$,then the $x$-coordinate of $Q$ is equal to

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