The distance of two points P and Q on the parabola `y^(2) = 4ax` from the focus S are 3 and 12 respectively. The distance of the point of intersection of the tangents at P and Q from the focus S is

Options

A. 8
B. 6
C. 9
D. 12

Explanation

To solve the problem, we need to find the distance of the point of intersection of the tangents at points P and Q on the parabola \( y^2 = 4ax \) from the focus S. The distances from the focus to points P and Q are given as 3 and 12, respectively.

### Step-by-Step Solution:

1. **Understand the Parabola**:
The equation of the parabola is given as \( y^2 = 4ax \). The focus of this parabola is at the point \( S(a, 0) \).

2. **Identify Points P and Q**:
Let the points P and Q on the parabola be represented in terms of parameters \( t_1 \) and \( t_2 \):
- Point \( P \) is \( (at_1^2, 2at_1) \)
- Point \( Q \) is \( (at_2^2, 2at_2) \)

3. **Calculate Distances from Focus S**:
The distances from the focus \( S(a, 0) \) to points P and Q are given as:
- \( SP = \sqrt{(at_1^2 - a)^2 + (2at_1 - 0)^2} \)
- \( SQ = \sqrt{(at_2^2 - a)^2 + (2at_2 - 0)^2} \)

According to the problem:
- \( SP = 3 \)
- \( SQ = 12 \)

4. **Use the Property of Parabolas**:
For points on the parabola, the distances can be expressed as:
- \( SP = \sqrt{(a(t_1^2 - 1))^2 + (2at_1)^2} = \sqrt{a^2(t_1^2 - 1)^2 + 4a^2t_1^2} = a\sqrt{(t_1^2 - 1)^2 + 4t_1^2} \)
- \( SQ = \sqrt{(a(t_2^2 - 1))^2 + (2at_2)^2} = a\sqrt{(t_2^2 - 1)^2 + 4t_2^2} \)

5. **Set Up the Equations**:
From the distances:
- \( a\sqrt{(t_1^2 - 1)^2 + 4t_1^2} = 3 \)
- \( a\sqrt{(t_2^2 - 1)^2 + 4t_2^2} = 12 \)

6. **Use the Tangent Intersection Point Formula**:
The coordinates of the point of intersection of the tangents at points P and Q is given by:
\[ T\left(a t_1 t_2, a(t_1 + t_2)\right) \]

7. **Calculate the Distance ST**:
The distance \( ST \) from the focus \( S(a, 0) \) to the point \( T \) is:
\[ ST = \sqrt{(at_1 t_2 - a)^2 + (a(t_1 + t_2) - 0)^2} \]
Simplifying gives:
\[ ST = a\sqrt{(t_1 t_2 - 1)^2 + (t_1 + t_2)^2} \]

8. **Using the Property of Distances**:
By the property of parabolas, we know:
\[ ST^2 = SP \cdot SQ \]
Therefore:
\[ ST^2 = 3 \cdot 12 = 36 \]
Thus:
\[ ST = \sqrt{36} = 6 \]

### Final Answer:
The distance of the point of intersection of the tangents at P and Q from the focus S is **6**.