Two mutually perpendicular tangents of the parabola `y^(2)=4ax` at the points `Q_(1) and Q_(2)` on it meet its axis in `P_(1)` and `P_(2)`. If S is the focus of the parabola, then the value of `((1)/(SP_(1))+(1)/(SP_(2)))^(-1)` is equal

Options

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

Answer

To solve the problem, we need to find the value of \(\left(\frac{1}{SP_1} + \frac{1}{SP_2}\right)^{-1}\) where \(S\) is the focus of the parabola \(y^2 = 4ax\), and \(P_1\) and \(P_2\) are the points where the tangents at points \(Q_1\) and \(Q_2\) meet the axis of the parabola.

### Step-by-Step Solution:

1. **Identify the Focus of the Parabola:**
The parabola \(y^2 = 4ax\) has its focus at the point \(S(a, 0)\).

2. **Equation of Tangents:**
The equation of the tangent to the parabola at point \(Q(t)\) is given by:
\[ y = tx + at^2 \]
For points \(Q_1\) and \(Q_2\) with parameters \(t_1\) and \(t_2\), the equations of the tangents are:
\[ y = t_1x + at_1^2 \quad \text{(for } Q_1\text{)} \]
\[ y = t_2x + at_2^2 \quad \text{(for } Q_2\text{)} \]

3. **Finding Points of Intersection with the Axis:**
The tangents meet the x-axis where \(y = 0\).
For \(P_1\) (intersection of tangent at \(Q_1\)):
\[ 0 = t_1x + at_1^2 \implies x = -\frac{at_1^2}{t_1} = -at_1 \]
Thus, \(P_1(-at_1, 0)\).

For \(P_2\) (intersection of tangent at \(Q_2\)):
\[ 0 = t_2x + at_2^2 \implies x = -\frac{at_2^2}{t_2} = -at_2 \]
Thus, \(P_2(-at_2, 0)\).

4. **Distance from Focus to Points:**
The distances \(SP_1\) and \(SP_2\) can be calculated as follows:
\[ SP_1 = \sqrt{((-at_1) - a)^2 + (0 - 0)^2} = \sqrt{(a(-t_1 - 1))^2} = a(t_1 + 1) \]
\[ SP_2 = \sqrt{((-at_2) - a)^2 + (0 - 0)^2} = \sqrt{(a(-t_2 - 1))^2} = a(t_2 + 1) \]

5. **Finding \(\frac{1}{SP_1} + \frac{1}{SP_2}\):**
Now we can find:
\[ \frac{1}{SP_1} = \frac{1}{a(t_1 + 1)}, \quad \frac{1}{SP_2} = \frac{1}{a(t_2 + 1)} \]
Therefore,
\[ \frac{1}{SP_1} + \frac{1}{SP_2} = \frac{1}{a(t_1 + 1)} + \frac{1}{a(t_2 + 1)} = \frac{(t_2 + 1) + (t_1 + 1)}{a(t_1 + 1)(t_2 + 1)} = \frac{t_1 + t_2 + 2}{a(t_1 + 1)(t_2 + 1)} \]

6. **Finding the Inverse:**
The value we need is:
\[ \left(\frac{1}{SP_1} + \frac{1}{SP_2}\right)^{-1} = \frac{a(t_1 + 1)(t_2 + 1)}{t_1 + t_2 + 2} \]

### Conclusion:
The final answer is:
\[ \left(\frac{1}{SP_1} + \frac{1}{SP_2}\right)^{-1} = a \]