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Q.
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
$ Q =\left( t , t ^2\right) $
$ m _{ CQ }= m _{\text {normal }} $
$ \frac{ t ^2+1}{ t -1}=-\frac{1}{2 t }$
Let $ f ( t )=2 t ^3+3 t -1 $
$ f \left(\frac{1}{4}\right) f \left(\frac{1}{3}\right)<0 \Rightarrow t \in\left(\frac{1}{4}, \frac{1}{3}\right) $
$P \equiv(1+\cos (90+\theta),-1+\sin (90+\theta))$
$ P =(1-\sin \theta,-1+\cos \theta)$
$ m _{\text {normal }}= m _{ CP } \Rightarrow-\frac{1}{2 t }=\frac{\cos \theta}{-\sin \theta} \Rightarrow \tan \theta=2 t$
$x =1-\sin \theta=1-\frac{2 t }{\sqrt{1+4 t ^2}}= g ( t ) \text { (let) }$
$\Rightarrow g ^{\prime}( t )<0$
$ g ( t ) \downarrow \text { function } $
$ t \in\left(\frac{1}{4}, \frac{1}{3}\right)$
$ \Rightarrow g ( t ) \in(0.44,0.485) \in\left(\frac{1}{4}, \frac{1}{2}\right)$
