JEE Main 2022 · 26 Jul · Shift 2 · Mathematics Q5

← JEE Main 2022 · 26 Jul · Shift 2

MathematicsSingle correct+4 / -1Question 5 of 90

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

\left(0,\dfrac14\right)

\left(\dfrac12,\dfrac34\right)

\left(\dfrac14,\dfrac12\right)

\left(\dfrac34,1\right)

Answer and worked solution

Official answer: C

Common normal. The shortest PQ lies along the normal to $y=x^2$ at $Q(t,t^2)$ that passes through the centre $(1,-1)$ : $(t-1)+2t(t^2+1)=0$ , i.e. $2t^3+3t-1=0$ , so $t\approx0.313$ and $Q\approx(0.313,0.098)$ .

P is where this segment meets the circle: the centre is $1.295$ from Q, so P's abscissa is $1-\frac{0.687}{1.295}\approx0.47$ .

Interval $\left(\frac14,\frac12\right)$ . (C)

Topics:Maxima and minima Parabola Equation of a circle

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JEE Main 2022 · 26 Jul · Shift 2, Mathematics question 5