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(C) The shortest distance between a line and a curve occurs at a point on the curve where the tangent is parallel to the given line.The given line is

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$\frac{7}{4\sqrt{2}}$

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(C) The shortest distance between a line and a curve occurs at a point on the curve where the tangent is parallel to the given line.The given line is $y = x$,which has a slope of $m = 1$.The curve is $y^2 = x - 2$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 1$,so $\frac{dy}{dx} = \frac{1}{2y}$.Setting the slope of the tangent equal to the slope of the line: $\frac{1}{2y} = 1 \Rightarrow y = \frac{1}{2}$.Substituting $y = \frac{1}{2}$ into the curve equation: $(\frac{1}{2})^2 = x - 2 \Rightarrow \frac{1}{4} = x - 2 \Rightarrow x = 2 + \frac{1}{4} = \frac{9}{4}$.So,the point on the curve is $P(\frac{9}{4}, \frac{1}{2})$.The shortest distance is the perpendicular distance from point $P(\frac{9}{4}, \frac{1}{2})$ to the line $x - y = 0$.Distance $d = \frac{|x_1 - y_1|}{\sqrt{1^2 + (-1)^2}} = \frac{|\frac{9}{4} - \frac{1}{2}|}{\sqrt{2}} = \frac{|\frac{9-2}{4}|}{\sqrt{2}} = \frac{7}{4\sqrt{2}}$.

The shortest distance between the line $y = x$ and the curve $y^2 = x - 2$ is

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