The interval in which abscissa of point P on `y=x^2` lies such that its distance from `(x-1)^2+(y+1)^2=1` is minimum is

Options

A. `0 lt x lt 1/4`
B. `1/4 lt x lt 1/2`
C. `1/2 lt x lt 3/4`
D. `3/4 lt x lt 1`

Answer

To solve the problem, we need to find the interval in which the abscissa (x-coordinate) of point P on the parabola \(y = x^2\) lies such that its distance from the circle defined by \((x - 1)^2 + (y + 1)^2 = 1\) is minimized.

### Step-by-Step Solution:

1. **Identify the point P on the parabola**:
The point P can be represented as \(P(x, y) = (x, x^2)\).

2. **Identify the center and radius of the circle**:
The equation of the circle is \((x - 1)^2 + (y + 1)^2 = 1\).
- Center of the circle \(C = (1, -1)\)
- Radius \(r = 1\)

3. **Calculate the distance from point P to the center of the circle**:
The distance \(D\) from point P to the center C is given by:
\[ D = \sqrt{(x - 1)^2 + (x^2 + 1)^2} \]

4. **Express the distance from the circle**:
The distance from point P to the circle is:
\[ d = D - r = \sqrt{(x - 1)^2 + (x^2 + 1)^2} - 1 \]

5. **Minimize the distance**:
To minimize \(d\), we can minimize \(D\) since \(r\) is constant. Thus, we need to minimize:
\[ f(x) = (x - 1)^2 + (x^2 + 1)^2 \]

6. **Differentiate the function**:
Differentiate \(f(x)\) with respect to \(x\):
\[ f'(x) = 2(x - 1) + 2(x^2 + 1)(2x) \]
Simplifying gives:
\[ f'(x) = 2(x - 1) + 4x(x^2 + 1) \]

7. **Set the derivative to zero**:
Set \(f'(x) = 0\):
\[ 2(x - 1) + 4x(x^2 + 1) = 0 \]
Simplifying further leads to:
\[ 2x^3 + 3x - 1 = 0 \]

8. **Find the roots of the equation**:
We can use numerical methods or graphing to find the roots of \(2x^3 + 3x - 1 = 0\).

9. **Test intervals**:
We can test intervals around the roots to find where the derivative changes sign. This will help us determine the intervals where the distance is minimized.

10. **Determine the interval**:
After testing values in the intervals, we find that the minimum distance occurs in the interval:
\[ \left(\frac{1}{4}, \frac{1}{2}\right) \]

### Final Answer:
The interval in which the abscissa of point P lies such that its distance from the circle is minimized is:
\[ \left(\frac{1}{4}, \frac{1}{2}\right) \]