JEE Main 2022 (Online) 26th July Evening ShiftQuestion 1
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
Differentiation and its Application
Solution
Concept. The shortest distance from a point $Q$ to the circle is along $QC$ ($C$ = centre), and equals $QC-1$. So minimise $QC$ over the parabola; $P$ is where $CQ$ meets the circle. Formulas used. (1) $QC^2=(t-1)^2+(t^2+1)^2$ for $Q=(t,t^2)$(2) $P=C+\dfrac{Q-C}{QC}$ (unit step from $C$ towards $Q$)Step 1 - Minimise. By (1), $$\frac{d}{dt}QC^2=2(t-1)+4t(t^2+1)=0\ \Longrightarrow\ 2t^3+3t-1=0.$$ This cubic is increasing, with one root $t\approx0.313$. Step 2 - $QC$. $QC^2\approx(0.687)^2+(1.098)^2$, so $QC\approx1.295$. Step 3 - $P$. By (2), the abscissa of $P$ is $$1+\frac{t-1}{QC}\approx1-\frac{0.687}{1.295}\approx0.47.$$Step 4 - Interval. $0.47\in\left(\frac14,\frac12\right)$. Answer: (C)