Consider point A(6, 30), point B(24, 6) and line AB: 4x+3y = 114.
Point `P(0, lambda)` is a point on y-axis such that `0 lt lambda lt 38 " and point " Q(0, lambda)` is a point on y-axis such that `lambda gt 38`.
For all positions of pont P, angle APB is maximum when point P is

Options

A. (0, 12)
B. (0, 15)
C. (0, 18)
D. (0, 21)

Explanation

To solve the problem, we need to find the point P on the y-axis such that the angle APB is maximized. We will use the cosine rule and the distance formula to derive the necessary equations.

### Step-by-Step Solution:

1. **Identify Points**:
- Let point A be \( A(6, 30) \) and point B be \( B(24, 6) \).
- Point P is represented as \( P(0, \lambda) \) where \( 0 < \lambda < 38 \).

2. **Calculate Distances**:
- We need to find the distances AP and BP using the distance formula:
\[ AP = \sqrt{(6 - 0)^2 + (30 - \lambda)^2} = \sqrt{36 + (30 - \lambda)^2} \]
\[ BP = \sqrt{(24 - 0)^2 + (6 - \lambda)^2} = \sqrt{576 + (6 - \lambda)^2} \]

3. **Expand the Distances**:
- For \( AP \):
\[ AP = \sqrt{36 + (30 - \lambda)^2} = \sqrt{36 + 900 - 60\lambda + \lambda^2} = \sqrt{\lambda^2 - 60\lambda + 936} \]
- For \( BP \):
\[ BP = \sqrt{576 + (6 - \lambda)^2} = \sqrt{576 + 36 - 12\lambda + \lambda^2} = \sqrt{\lambda^2 - 12\lambda + 612} \]

4. **Calculate Distance AB**:
- Using the distance formula for points A and B:
\[ AB = \sqrt{(24 - 6)^2 + (6 - 30)^2} = \sqrt{18^2 + (-24)^2} = \sqrt{324 + 576} = \sqrt{900} = 30 \]

5. **Apply the Cosine Rule**:
- According to the cosine rule:
\[ \cos(\angle APB) = \frac{AP^2 + BP^2 - AB^2}{2 \cdot AP \cdot BP} \]
- Substitute the values:
\[ \cos(\angle APB) = \frac{(\lambda^2 - 60\lambda + 936) + (\lambda^2 - 12\lambda + 612) - 900}{2 \cdot AP \cdot BP} \]
- Simplifying the numerator:
\[ = \frac{2\lambda^2 - 72\lambda + 648}{2 \cdot AP \cdot BP} \]

6. **Maximize the Angle**:
- The angle APB is maximum when \( \cos(\angle APB) = 0 \), which occurs when \( \angle APB = 90^\circ \).
- Set the numerator equal to zero:
\[ 2\lambda^2 - 72\lambda + 648 = 0 \]
- Simplifying gives:
\[ \lambda^2 - 36\lambda + 324 = 0 \]
- Factoring or using the quadratic formula:
\[ (\lambda - 18)^2 = 0 \implies \lambda = 18 \]

7. **Conclusion**:
- The point P that maximizes the angle APB is:
\[ P(0, 18) \]