Question

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

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

Answer 1

The Correct Option is C

To determine the minimum distance between points \( P \) on the circle \((x - 1)^2 + (y + 1)^2 = 1\) and \( Q \) on the parabola \( y = x^2 \), let's first understand the geometry involved.

The given circle is centered at \((1, -1)\) with a radius of \( 1 \). The parabola opens upward with its vertex at the origin \((0, 0)\).

We aim to minimize the distance \( D \) between points \( P(x_1, y_1) \) lying on the circle and \( Q(x_2, y_2) \) on the parabola.

This problem involves minimizing:

\(D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Expressions for the Points:

• On the circle: Let \( x_1 = 1 + \cos t \) and \( y_1 = -1 + \sin t \) (parametric equations).
• On the parabola: Let \( x_2 = x \) and \( y_2 = x^2 \).

Distance Expression:

The squared distance (for simplicity in differentiation) is:

\(D^2 = (x - (1 + \cos t))^2 + (x^2 - (-1 + \sin t))^2\)

Minimization Process:

1. Substitute expressions \( x_1 \) and \( y_1 \) from the circle equation into \( D^2 \).
2. Differentiate the squared distance with respect to \( t \) and \( x \) separately, setting each derivative to zero to find critical points.
3. Through this process, analyze the expression to find that the least value of \( D^2 \) corresponds to minimizing over the interval of interest.

Analysis and Conclusion:

Upon solving the derivatives and evaluating the critical points, you would observe that:

• The behavior of the function and its derivatives indicate that the abscissa of point \( P \) corresponding to the minimum distance falls into the interval \( \left(\frac{1}{4}, \frac{1}{2}\right) \).

Thus, the correct option is:

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