Tangents to the parabola `y^2=4ax` at `P(at_1^2,2at_1)and Q(at_2^2,2at_2)` meet at T. If `DeltaPTQ` is right - angled at T, then `1/(PS)+1/(QS)` is equal to (where , S is the focus of the given parabola)

Options

A. `1/a`
B. `2/a`
C. `1/(2a)`
D. `1/(4a)`

Answer

To solve the problem, we need to find the value of \( \frac{1}{PS} + \frac{1}{QS} \) given that the tangents to the parabola \( y^2 = 4ax \) at points \( P(at_1^2, 2at_1) \) and \( Q(at_2^2, 2at_2) \) meet at point \( T \) and triangle \( PTQ \) is right-angled at \( T \).

### Step-by-Step Solution:

1. **Identify the Focus and Points**:
The focus \( S \) of the parabola \( y^2 = 4ax \) is at \( (a, 0) \). The points \( P \) and \( Q \) are given as:
\[ P(at_1^2, 2at_1) \quad \text{and} \quad Q(at_2^2, 2at_2) \]

2. **Find the Tangents at Points P and Q**:
The equation of the tangent to the parabola at point \( P \) is given by:
\[ yy_1 = 2a(x + x_1) \]
Substituting \( P \):
\[ y(2at_1) = 2a(x + at_1^2) \implies yt_1 = x + at_1^2 \quad \text{(1)} \]
Similarly, for point \( Q \):
\[ y(2at_2) = 2a(x + at_2^2) \implies yt_2 = x + at_2^2 \quad \text{(2)} \]

3. **Find Point T**:
The point \( T \) is the intersection of the two tangents (1) and (2). To find \( T \), we can solve these two equations simultaneously.

4. **Use the Right-Angle Condition**:
Given that triangle \( PTQ \) is right-angled at \( T \), we can use the property of the slopes of the tangents:
\[ m_1 \cdot m_2 = -1 \]
where \( m_1 \) and \( m_2 \) are the slopes of the tangents at points \( P \) and \( Q \) respectively. The slopes can be derived from the tangent equations.

5. **Calculate Distances PS and QS**:
The distances \( PS \) and \( QS \) can be calculated using the distance formula:
\[ PS = \sqrt{(at_1^2 - a)^2 + (2at_1 - 0)^2} \]
\[ QS = \sqrt{(at_2^2 - a)^2 + (2at_2 - 0)^2} \]

6. **Find \( \frac{1}{PS} + \frac{1}{QS} \)**:
Using the distances calculated, we can find:
\[ \frac{1}{PS} + \frac{1}{QS} \]
Substitute the values of \( PS \) and \( QS \) into this expression.

7. **Use the Relationship Between \( t_1 \) and \( t_2 \)**:
Since \( t_1 t_2 = -1 \), we can express \( t_2 \) in terms of \( t_1 \) as \( t_2 = -\frac{1}{t_1} \). Substitute this into the expression for \( \frac{1}{PS} + \frac{1}{QS} \) to simplify.

8. **Final Calculation**:
After simplification, we find:
\[ \frac{1}{PS} + \frac{1}{QS} = \frac{1}{a} \]

### Conclusion:
Thus, the final result is:
\[ \frac{1}{PS} + \frac{1}{QS} = \frac{1}{a} \]