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(C) Given the curve is $2 x^2+y^2=2 x$. Rearranging terms: $2 x^2-2 x+y^2=0$. Completing the square: $2(x^2-x)+y^2=0 \Rightarrow 2(x-\frac{1}{2})^2+y^

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$\sqrt{1-2 a+2 a^2}$

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(C) Given the curve is $2 x^2+y^2=2 x$. Rearranging terms: $2 x^2-2 x+y^2=0$. Completing the square: $2(x^2-x)+y^2=0 \Rightarrow 2(x-\frac{1}{2})^2+y^2=\frac{1}{2}$. Dividing by $\frac{1}{2}$: $\frac{(x-\frac{1}{2})^2}{1/4} + \frac{y^2}{1/2} = 1$. This is an ellipse with center $(\frac{1}{2}, 0)$,$a^2 = \frac{1}{4} \Rightarrow a = \frac{1}{2}$,and $b^2 = \frac{1}{2} \Rightarrow b = \frac{1}{\sqrt{2}}$. Let a point $P$ on the ellipse be $(\frac{1}{2} + \frac{1}{2} \cos 	heta, \frac{1}{\sqrt{2}} \sin 	heta)$. The distance $PQ$ from $Q(a, 0)$ is $PQ^2 = (\frac{1}{2} + \frac{1}{2} \cos 	heta - a)^2 + (\frac{1}{\sqrt{2}} \sin 	heta)^2$. $PQ^2 = (\frac{1}{2}-a)^2 + \frac{1}{4} \cos^2 	heta + (\frac{1}{2}-a) \cos 	heta + \frac{1}{2} \sin^2 	heta$. Substituting $\sin^2 	heta = 1 - \cos^2 	heta$: $PQ^2 = (\frac{1}{2}-a)^2 + \frac{1}{4} \cos^2 	heta + (\frac{1}{2}-a) \cos 	heta + \frac{1}{2} - \frac{1}{2} \cos^2 	heta$. $PQ^2 = -\frac{1}{4} \cos^2 	heta + (\frac{1}{2}-a) \cos 	heta + (\frac{1}{2}-a)^2 + \frac{1}{2}$. This is a quadratic in $\cos 	heta$. The maximum occurs at $\cos 	heta = \frac{-b}{2A} = \frac{-(1/2-a)}{2(-1/4)} = 1-2a$. Substituting $\cos 	heta = 1-2a$ into the expression for $PQ^2$: $PQ^2 = -\frac{1}{4}(1-2a)^2 + (\frac{1}{2}-a)(1-2a) + \frac{1}{4} - a + a^2 + \frac{1}{2}$. $PQ^2 = -\frac{1}{4}(1-4a+4a^2) + (\frac{1}{2}-a-a+2a^2) + \frac{3}{4} - a + a^2$. $PQ^2 = -\frac{1}{4} + a - a^2 + \frac{1}{2} - 2a + 2a^2 + \frac{3}{4} - a + a^2 = 2a^2 - 2a + 1$. Thus,the longest distance is $\sqrt{2a^2 - 2a + 1}$.

The longest distance of the point $(a, 0)$ from the curve $2 x^2+y^2=2 x$ is

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