Let P and Q be any points on the curves (x-1)²+(y+1)²=1 and y=x², respectively. The distance between P and Q is minimum for some value of the abscissa of P in the interval :

Solution

Maths sir · 5 steps

Forget P for a moment: find the parabola point nearest the circle's centre, then walk 1 unit back toward it.

1. Use the centre. The circle has centre C(1, -1) and radius 1. For any point Q outside the circle, the nearest point P on the circle lies on segment CQ, and PQ = CQ - 1. So minimising PQ means minimising CQ.
2. Distance to parabola. Take Q = (t, t²). Then CQ² = f(t) = (t-1)² + (t²+1)². Set its derivative to zero.

   f'(t) = 2(t-1) + 4t(t²+1) = 0 ⇒ 2t³ + 3t - 1 = 0

3. Locate the root. At t = 0.3 the cubic is -0.046, and at t = 0.32 it is +0.026, so t ≈ 0.313. This is the only root, because 6t² + 3 > 0 means the cubic is always increasing. So Q ≈ (0.313, 0.098) and CQ ≈ √(0.472 + 1.205) ≈ 1.295, which is more than 1. The parabola does lie outside the circle.
4. Find P. P is 1 unit from C towards Q, so x_P = 1 + (t - 1)/CQ ≈ 1 - 0.687/1.295 ≈ 1 - 0.530 = 0.470.
5. So. The abscissa of P is about 0.47, which lies in (1/4, 1/2). The answer is (c).

The idea to keep. For the shortest distance from a circle to a curve, minimise the distance from the circle's centre to the curve, then subtract the radius. The shortest path always runs along a normal through the centre.

Answer from the NTA final answer key. Solution by PrepWiser.