To solve the problem step by step, we will follow the outlined approach:
### Step 1: Identify the foci of the ellipse
The given ellipse is \(\frac{x^2}{9} + \frac{y^2}{8} = 1\). The foci of an ellipse are given by the formula \(c = \sqrt{a^2 - b^2}\), where \(a^2 = 9\) and \(b^2 = 8\).
- Calculate \(c\):
\[
c = \sqrt{9 - 8} = \sqrt{1} = 1
\]
Thus, the foci are at \(F_1(-1, 0)\) and \(F_2(1, 0)\).
### Step 2: Find the equation of the parabola
The parabola has its vertex at the origin \((0,0)\) and focus at \(F_2(1,0)\). The standard form of a parabola that opens to the right is given by:
\[
y^2 = 4px
\]
where \(p\) is the distance from the vertex to the focus. Here, \(p = 1\), so the equation of the parabola is:
\[
y^2 = 4x
\]
### Step 3: Find the points of intersection \(M\) and \(N\)
To find the intersection points of the ellipse and the parabola, substitute \(y^2 = 4x\) into the ellipse equation:
\[
\frac{x^2}{9} + \frac{4x}{8} = 1
\]
This simplifies to:
\[
\frac{x^2}{9} + \frac{x}{2} - 1 = 0
\]
Multiply through by 18 to eliminate the denominators:
\[
2x^2 + 9x - 18 = 0
\]
Now, use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- Here, \(a = 2\), \(b = 9\), and \(c = -18\).
- Calculate the discriminant:
\[
D = 9^2 - 4 \cdot 2 \cdot (-18) = 81 + 144 = 225
\]
- Now, find the roots:
\[
x = \frac{-9 \pm 15}{4}
\]
This gives:
\[
x_1 = \frac{6}{4} = \frac{3}{2}, \quad x_2 = \frac{-24}{4} = -6
\]
Since \(M\) is in the first quadrant and \(N\) in the fourth quadrant, we take:
\[
M\left(\frac{3}{2}, \sqrt{6}\right), \quad N\left(-6, -\sqrt{24}\right)
\]
### Step 4: Find the coordinates of point \(R\)
The tangents at points \(M\) and \(N\) can be found using the tangent line equation for the ellipse:
\[
\frac{x x_1}{9} + \frac{y y_1}{8} = 1
\]
For point \(M\):
\[
\frac{x \cdot \frac{3}{2}}{9} + \frac{y \cdot \sqrt{6}}{8} = 1
\]
Set \(y = 0\) to find \(R\):
\[
\frac{x \cdot \frac{3}{2}}{9} = 1 \implies x = 6 \implies R(6, 0)
\]
### Step 5: Find the coordinates of point \(Q\)
The normal to the parabola at point \(M\) can be found using the normal line equation:
\[
y - y_1 = -\frac{1}{2} (x - x_1)
\]
Substituting \(M\):
\[
y - \sqrt{6} = -\frac{1}{2} \left(x - \frac{3}{2}\right)
\]
Set \(y = 0\) to find \(Q\):
\[
0 - \sqrt{6} = -\frac{1}{2} \left(x - \frac{3}{2}\right)
\]
Solving gives:
\[
x = \frac{7}{2} \implies Q\left(\frac{7}{2}, 0\right)
\]
### Step 6: Calculate the areas
1. **Area of triangle \(MQR\)**:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \left(6 - \frac{7}{2}\right) \times \sqrt{6} = \frac{1}{2} \times \frac{5}{2} \times \sqrt{6} = \frac{5\sqrt{6}}{4}
\]
2. **Area of quadrilateral \(MF_1NF_2\)**:
This can be calculated as twice the area of triangle \(MF_1N\):
\[
\text{Area} = 2 \times \frac{1}{2} \times \text{base} \times \text{height} = 2 \times \frac{1}{2} \times \left(-1 - \frac{3}{2}\right) \times \sqrt{6} = 2 \sqrt{6}
\]
### Step 7: Find the ratio of the areas
The ratio of the area of triangle \(MQR\) to the area of quadrilateral \(MF_1NF_2\) is:
\[
\text{Ratio} = \frac{\frac{5\sqrt{6}}{4}}{2\sqrt{6}} = \frac{5}{8}
\]
Thus, the final answer is:
\[
\text{The ratio of the area of triangle } MQR \text{ to the area of quadrilateral } MF_1NF_2 \text{ is } 5:8.
\]
