To solve the given problem step by step, we will analyze the information provided and derive the necessary equations.
### Step 1: Understand the given parabola and its properties
The parabola is given by the equation \( y^2 = 4x \). This is a standard form of a parabola that opens to the right, where \( A = 1 \).
**Hint:** Recall that the general form of a parabola is \( y^2 = 4Ax \).
### Step 2: Write the equation of the normal to the parabola
The slope of the normal to the parabola at a point \( (x_0, y_0) \) is given by:
\[
y = mx - 2am - am^2
\]
where \( m \) is the slope of the tangent at that point, and \( a \) is the parameter of the parabola.
For \( y^2 = 4x \), the slope \( m \) can be derived from the point on the parabola. The coordinates of the point on the parabola can be expressed as \( (t^2, 2t) \) for some parameter \( t \).
**Hint:** Remember that the slope of the tangent line at point \( (t^2, 2t) \) is \( \frac{dy}{dx} = \frac{2}{t} \).
### Step 3: Set up the equation for the normals at points A, B, and C
Given that the normals at points \( A, B, \) and \( C \) are concurrent at the point \( (9, 6) \), we can substitute \( x = 9 \) and \( y = 6 \) into the normal equation.
The normal at point \( (t^2, 2t) \) becomes:
\[
6 = m \cdot 9 - 2 \cdot 1 \cdot m - m^2
\]
This simplifies to:
\[
m^2 - 7m + 6 = 0
\]
**Hint:** This is a quadratic equation in \( m \). Use the quadratic formula to find the roots.
### Step 4: Solve the quadratic equation
Using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
m = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}
\]
\[
m = \frac{7 \pm \sqrt{49 - 24}}{2}
\]
\[
m = \frac{7 \pm \sqrt{25}}{2}
\]
\[
m = \frac{7 \pm 5}{2}
\]
Thus, the roots are:
\[
m_1 = 6, \quad m_2 = 1
\]
And we can also find another root by checking for \( m = -3 \).
**Hint:** Check the values of \( m \) to confirm they satisfy the original cubic equation.
### Step 5: Calculate the center of the circle
The center of the circle \( S \) is given by \( (\alpha, \beta) \). The average of the slopes \( m_1, m_2, m_3 \) gives us the coordinates of the center:
\[
\alpha = \frac{m_1 + m_2 + m_3}{3}
\]
Substituting the values:
\[
\alpha = \frac{6 + 1 - 3}{3} = \frac{4}{3}
\]
**Hint:** The center of the circle can also be derived from the intersection properties of the normals.
### Step 6: Find \( \alpha + \beta \)
From the earlier calculations, we can find \( \beta \) using the coordinates of the center:
\[
\beta = \frac{3}{2}
\]
Thus:
\[
\alpha + \beta = \frac{4}{3} + \frac{3}{2}
\]
Finding a common denominator (which is 6):
\[
\alpha + \beta = \frac{8}{6} + \frac{9}{6} = \frac{17}{6}
\]
**Hint:** Ensure you convert to a common denominator when adding fractions.
### Conclusion
After performing all calculations, we find that the sum \( \alpha + \beta \) is \( \frac{17}{6} \).
### Final Answer
The answer is \( \alpha + \beta = \frac{17}{6} \).
