The longest distance of the point (a, 0) from | vedclass

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(C) Given,the curve is $2x^2+y^2=2x$. Rearranging the terms,we get $2x^2-2x+y^2=0$. Completing the square for $x$: $2(x^2-x)+y^2=0 \Rightarrow 2(x-\fr

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

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(C) Given,the curve is $2x^2+y^2=2x$. Rearranging the terms,we get $2x^2-2x+y^2=0$. Completing the square for $x$: $2(x^2-x)+y^2=0 \Rightarrow 2(x-\frac{1}{2})^2+y^2=\frac{1}{2}$. Dividing by $\frac{1}{2}$,we get $\frac{(x-\frac{1}{2})^2}{1/4} + \frac{y^2}{1/2} = 1$,which represents an ellipse. Let a point $P$ on the ellipse be $P(\frac{1}{2}+\frac{1}{2}\cos	heta, \frac{1}{\sqrt{2}}\sin	heta)$. The distance $PQ$ from $Q(a, 0)$ is given by $PQ^2 = (\frac{1}{2}+\frac{1}{2}\cos	heta-a)^2 + (\frac{1}{\sqrt{2}}\sin	heta)^2$. $PQ^2 = ((\frac{1}{2}-a)+\frac{1}{2}\cos	heta)^2 + \frac{1}{2}\sin^2	heta$. $PQ^2 = (\frac{1}{2}-a)^2 + \frac{1}{4}\cos^2	heta + (\frac{1}{2}-a)\cos	heta + \frac{1}{2}(1-\cos^2	heta)$. $PQ^2 = \frac{1}{4}-a+a^2 + \frac{1}{4}\cos^2	heta + (\frac{1}{2}-a)\cos	heta + \frac{1}{2} - \frac{1}{2}\cos^2	heta$. $PQ^2 = a^2-a+\frac{3}{4} - \frac{1}{4}\cos^2	heta + (\frac{1}{2}-a)\cos	heta$. To find the maximum distance,we differentiate with respect to $\cos	heta$ and set to $0$,or analyze the quadratic in $\cos	heta$. After simplification,the maximum distance is $\sqrt{1-2a+2a^2}$.

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

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