Let P and Q be any points on the curves (x-1) | vedclass

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(C) The circle is $(x-1)^{2}+(y+1)^{2}=1$ with center $C(1, -1)$ and radius $r=1$. Let $Q(t, t^{2})$ be a point on the parabola $y=x^{2}$.The distance

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$\left(\frac{1}{4}, \frac{1}{2}\right)$

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(C) The circle is $(x-1)^{2}+(y+1)^{2}=1$ with center $C(1, -1)$ and radius $r=1$. Let $Q(t, t^{2})$ be a point on the parabola $y=x^{2}$.The distance between $P$ and $Q$ is minimum when $Q$ lies on the normal to the circle passing through $C$. Thus,the normal to the parabola at $Q$ must pass through the center $C(1, -1)$.The slope of the tangent to $y=x^{2}$ at $Q(t, t^{2})$ is $m_{T} = 2t$. The slope of the normal is $m_{N} = -\frac{1}{2t}$.The slope of the line $CQ$ is $\frac{t^{2}-(-1)}{t-1} = \frac{t^{2}+1}{t-1}$.Equating the slopes: $\frac{t^{2}+1}{t-1} = -\frac{1}{2t} \Rightarrow 2t^{3}+2t = -t+1 \Rightarrow 2t^{3}+3t-1=0$.Let $f(t) = 2t^{3}+3t-1$. Since $f(0) = -1$ and $f(1) = 4$,the root $t$ lies in $(0, 1)$. Specifically,$f(1/4) = 2(1/64) + 3/4 - 1 = 1/32 - 1/4 = -7/32  0$. Thus,$t \in (1/4, 1/2)$.The point $P$ on the circle is the intersection of the line $CQ$ and the circle. The abscissa of $P$ is $x_P = 1 + \cos \phi$,where $\phi$ is the angle of the line $CQ$. Since $	an \phi = \frac{t^{2}+1}{t-1} = -\frac{1}{2t}$,we find $x_P$ lies in the interval $\left(\frac{1}{4}, \frac{1}{2}ight)$.

Let $P$ and $Q$ be any points on the curves $(x-1)^{2}+(y+1)^{2}=1$ and $y=x^{2}$,respectively. The distance between $P$ and $Q$ is minimum for some value of the abscissa of $P$ in the interval

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