Question

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

• A. $\left(0,\frac{1}{4}\right)$
• B. $\left(\frac{1}{2},\frac{3}{4}\right)$
• C. $\left(\frac{1}{4},\frac{1}{2}\right)$
• D. $\left(\frac{3}{4},1\right)$

Answer 1

Answer: (C)

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