JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Application of Derivatives

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

Answer: (C)

Solution

The minimum distance occurs when the line joining the center

C (1, - 1)

of the circle to a point

Q (t, t^{2})

on the parabola is perpendicular to the parabola at Q.

The slope of the tangent to

y = x^{2}

at Q is

2 t

, so the normal has slope

- \frac{1}{2 t}

.

The slope of CQ is

\frac{t ^{2} + 1}{t - 1}

.

Setting them equal:

\frac{t ^{2} + 1}{t - 1} = - \frac{1}{2 t}

, which gives

2 t^{3} + 3 t - 1 = 0

.

Let

f (t) = 2 t^{3} + 3 t - 1

. Testing:

f (1/4) < 0

and

f (1/3) > 0

, so the root lies in

(1/4, 1/3)

.

The corresponding x-coordinate of P (using the parametric form of the circle and the perpendicularity condition) lies in the interval

(\frac{1}{4}, \frac{1}{2})

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option