"Let $\mathrm{P}$ and $\mathrm{Q}$ 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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Accepted Answer

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$y = mx + 2a + {1 \over {{m^2}}}$ (Equation of normal to ${x^2} = 4ay$ in slope form) through $(1, - 1)$.

$4{m^3} + 6{m^2} + 1 = 0$

$\Rightarrow m \simeq - 1.6$

Slope of normal $\simeq {{ - 8} \over 5} = \tan \theta$

$$ \Rightarrow \cos \theta \simeq {{ - 5} \over {\sqrt {89} }},\,\sin \theta \simeq {8 \over {\sqrt {89} }}$$

$${x_p} = 1 + \cos \theta \simeq 1 - {5 \over {\sqrt {89} }} \in \left( {{1 \over 4},{1 \over 2}} \right)$$

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Let $\mathrm{P}$ and $\mathrm{Q}$ 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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Let $\mathrm{P}$ and $\mathrm{Q}$ 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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