P is a variable point on the curve y= f(x) and A is a fixed point in the plane not lying on the curve. If `PA^(2)` is minimum, then the angle between PA and the tangent at P is

Options

A. `(pi)/(4)`
B. `(pi)/(3)`
C. `(pi)/(2)`
D. None of these

Answer

To solve the problem, we need to find the angle between the line segment \( PA \) (from point \( P \) on the curve \( y = f(x) \) to the fixed point \( A \)) and the tangent line at point \( P \) when the distance \( PA^2 \) is minimized.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let \( P(x, f(x)) \) be a point on the curve.
- Let \( A(a, b) \) be a fixed point in the plane.
- We need to minimize the distance \( PA \).

2. **Distance Formula**:
- The distance \( PA \) can be expressed as:
\[ PA = \sqrt{(x - a)^2 + (f(x) - b)^2} \]
- Therefore, the square of the distance is:
\[ PA^2 = (x - a)^2 + (f(x) - b)^2 \]

3. **Minimizing \( PA^2 \)**:
- To find the minimum distance, we differentiate \( PA^2 \) with respect to \( x \) and set the derivative to zero:
\[ \frac{d(PA^2)}{dx} = 2(x - a) + 2(f(x) - b)f'(x) = 0 \]
- Simplifying this gives:
\[ (x - a) + (f(x) - b)f'(x) = 0 \]

4. **Geometric Interpretation**:
- The condition for \( PA^2 \) to be minimized implies that the line segment \( PA \) is perpendicular to the tangent line at point \( P \).
- This is because the shortest distance from a point to a line occurs along the perpendicular.

5. **Finding the Angle**:
- Since \( PA \) is perpendicular to the tangent at \( P \), the angle \( \theta \) between \( PA \) and the tangent line is:
\[ \theta = 90^\circ \quad \text{or} \quad \theta = \frac{\pi}{2} \text{ radians} \]

### Conclusion:
Thus, the angle between the line segment \( PA \) and the tangent at point \( P \) is \( \frac{\pi}{2} \) radians.