To solve the problem, we need to find the points of intersection of the given circle and parabola, determine the points where the tangents intersect the x-axis, and then calculate the areas of the triangles formed.
### Step 1: Find the Points of Intersection
The equations of the circle and the parabola are:
1. Circle: \( x^2 + y^2 = 9 \)
2. Parabola: \( y^2 = 8x \)
Substituting \( y^2 = 8x \) into the circle's equation:
\[
x^2 + 8x = 9
\]
Rearranging gives:
\[
x^2 + 8x - 9 = 0
\]
Now, we can solve this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{64 + 36}}{2} = \frac{-8 \pm 10}{2}
\]
Calculating the roots:
\[
x_1 = 1 \quad \text{and} \quad x_2 = -9
\]
Now, substituting \( x = 1 \) back into the parabola's equation to find \( y \):
\[
y^2 = 8(1) = 8 \implies y = \pm 2\sqrt{2}
\]
Thus, the points of intersection are:
- \( P(1, 2\sqrt{2}) \) in the first quadrant
- \( Q(1, -2\sqrt{2}) \) in the fourth quadrant
### Step 2: Find the Tangents to the Circle
The equation of the tangent to the circle at point \( P(1, 2\sqrt{2}) \) is given by:
\[
x_1x + y_1y = 9 \implies 1 \cdot x + 2\sqrt{2} \cdot y = 9
\]
Rearranging gives:
\[
x + 2\sqrt{2}y - 9 = 0
\]
To find where this tangent intersects the x-axis (\( y = 0 \)):
\[
x - 9 = 0 \implies x = 9 \implies R(9, 0)
\]
For point \( Q(1, -2\sqrt{2}) \):
\[
1 \cdot x - 2\sqrt{2} \cdot y = 9 \implies x - 2\sqrt{2}y - 9 = 0
\]
Setting \( y = 0 \):
\[
x - 9 = 0 \implies x = 9 \implies R(9, 0)
\]
### Step 3: Find the Tangents to the Parabola
The equation of the tangent to the parabola at point \( P(1, 2\sqrt{2}) \) is:
\[
yy_1 = 4(x + x_1) \implies y(2\sqrt{2}) = 4(x + 1)
\]
Rearranging gives:
\[
2\sqrt{2}y - 4x - 4 = 0
\]
Setting \( y = 0 \):
\[
-4x - 4 = 0 \implies x = -1 \implies S(-1, 0)
\]
For point \( Q(1, -2\sqrt{2}) \):
\[
y(-2\sqrt{2}) = 4(x + 1) \implies -2\sqrt{2}y - 4x - 4 = 0
\]
Setting \( y = 0 \):
\[
-4x - 4 = 0 \implies x = -1 \implies S(-1, 0)
\]
### Step 4: Calculate the Areas of Triangles
**Area of Triangle \( PQS \):**
- Base \( PS = 2 \) (from \( P(1, 2\sqrt{2}) \) to \( S(-1, 0) \))
- Height = \( 2\sqrt{2} \)
Area:
\[
\text{Area}_{PQS} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2\sqrt{2} = 2\sqrt{2}
\]
**Area of Triangle \( PQR \):**
- Base \( PR = 8 \) (from \( P(1, 2\sqrt{2}) \) to \( R(9, 0) \))
- Height = \( 2\sqrt{2} \)
Area:
\[
\text{Area}_{PQR} = \frac{1}{2} \times 8 \times 2\sqrt{2} = 8\sqrt{2}
\]
### Step 5: Find the Ratio of Areas
The ratio of the areas of triangle \( PQS \) to triangle \( PQR \):
\[
\text{Ratio} = \frac{\text{Area}_{PQS}}{\text{Area}_{PQR}} = \frac{2\sqrt{2}}{8\sqrt{2}} = \frac{1}{4}
\]
### Final Answer
The ratio of the areas of triangle \( PQS \) and triangle \( PQR \) is \( \frac{1}{4} \).
