To solve the problem, we need to follow these steps:
### Step 1: Identify the Ellipse and its Parameters
The given ellipse is:
\[
\frac{x^2}{8} + \frac{y^2}{4} = 1
\]
From this, we can identify:
- \( a^2 = 8 \) ⇒ \( a = 2\sqrt{2} \)
- \( b^2 = 4 \) ⇒ \( b = 2 \)
### Step 2: Calculate the Eccentricity
The eccentricity \( e \) of an ellipse is given by the formula:
\[
e = \sqrt{1 - \frac{b^2}{a^2}}
\]
Substituting the values:
\[
e = \sqrt{1 - \frac{4}{8}} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}
\]
### Step 3: Find the Point \( P \) on the Ellipse
The coordinates of any point \( P \) on the ellipse can be expressed as:
\[
P = (a \cos \theta, b \sin \theta)
\]
Since \( P \) is in the second quadrant, we have:
\[
P = (2\sqrt{2} \cos \theta, 2 \sin \theta)
\]
where \( \cos \theta < 0 \) and \( \sin \theta > 0 \).
### Step 4: Equation of the Tangent at Point \( P \)
The equation of the tangent to the ellipse at point \( P \) is given by:
\[
\frac{x}{a \cos \theta} + \frac{y}{b \sin \theta} = 1
\]
Substituting \( a \) and \( b \):
\[
\frac{x}{2\sqrt{2} \cos \theta} + \frac{y}{2 \sin \theta} = 1
\]
Multiplying through by \( 2\sqrt{2} \sin \theta \):
\[
x \sin \theta + \sqrt{2} y \cos \theta = 2\sqrt{2} \sin \theta
\]
### Step 5: Find the Slope of the Tangent
Rearranging the tangent equation gives:
\[
\sqrt{2} y \cos \theta = 2\sqrt{2} \sin \theta - x \sin \theta
\]
The slope \( m_1 \) of the tangent line is:
\[
m_1 = -\frac{\sin \theta}{\sqrt{2} \cos \theta} = -\frac{\tan \theta}{\sqrt{2}}
\]
### Step 6: Find the Slope of the Given Line
The line \( x + 2y = 0 \) can be rearranged to find its slope:
\[
y = -\frac{1}{2}x \quad \Rightarrow \quad m_2 = -\frac{1}{2}
\]
### Step 7: Use the Perpendicular Condition
Since the tangents are perpendicular:
\[
m_1 \cdot m_2 = -1
\]
Substituting the slopes:
\[
\left(-\frac{\tan \theta}{\sqrt{2}}\right) \left(-\frac{1}{2}\right) = -1
\]
This simplifies to:
\[
\frac{\tan \theta}{2\sqrt{2}} = 1 \quad \Rightarrow \quad \tan \theta = 2\sqrt{2}
\]
### Step 8: Calculate \( \sin \theta \) and \( \cos \theta \)
Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
Let \( \sin \theta = 2\sqrt{2} \cos \theta \). Using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[
(2\sqrt{2} \cos \theta)^2 + \cos^2 \theta = 1 \quad \Rightarrow \quad 8 \cos^2 \theta + \cos^2 \theta = 1 \quad \Rightarrow \quad 9 \cos^2 \theta = 1
\]
Thus,
\[
\cos^2 \theta = \frac{1}{9} \quad \Rightarrow \quad \cos \theta = -\frac{1}{3} \quad (\text{since } \theta \text{ is in the second quadrant})
\]
Then,
\[
\sin^2 \theta = 1 - \frac{1}{9} = \frac{8}{9} \quad \Rightarrow \quad \sin \theta = \frac{2\sqrt{2}}{3}
\]
### Step 9: Find Coordinates of Point \( P \)
Substituting \( \cos \theta \) and \( \sin \theta \) into the coordinates of \( P \):
\[
P = \left(2\sqrt{2} \left(-\frac{1}{3}\right), 2 \left(\frac{2\sqrt{2}}{3}\right)\right) = \left(-\frac{2\sqrt{2}}{3}, \frac{4\sqrt{2}}{3}\right)
\]
### Step 10: Find the Foci \( S \) and \( S' \)
The foci of the ellipse are given by:
\[
S = (ae, 0) = \left(2\sqrt{2} \cdot \frac{1}{\sqrt{2}}, 0\right) = (2, 0)
\]
\[
S' = (-2, 0)
\]
### Step 11: Calculate the Area of Triangle \( SPS' \)
Using the formula for the area of a triangle with vertices at \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|
\]
Substituting the coordinates:
\[
\text{Area} = \frac{1}{2} \left| 2\left(\frac{4\sqrt{2}}{3} - 0\right) + (-2)\left(0 - \frac{4\sqrt{2}}{3}\right) + \left(-\frac{2\sqrt{2}}{3}\right)(0 - 0) \right|
\]
This simplifies to:
\[
= \frac{1}{2} \left| \frac{8\sqrt{2}}{3} + \frac{8\sqrt{2}}{3} \right| = \frac{1}{2} \cdot \frac{16\sqrt{2}}{3} = \frac{8\sqrt{2}}{3}
\]
### Step 12: Final Calculation
Now, we need to calculate:
\[
(5 - e^2) \cdot \text{Area}
\]
Where \( e^2 = \frac{1}{2} \):
\[
5 - e^2 = 5 - \frac{1}{2} = \frac{10}{2} - \frac{1}{2} = \frac{9}{2}
\]
Thus:
\[
(5 - e^2) \cdot \text{Area} = \frac{9}{2} \cdot \frac{8\sqrt{2}}{3} = \frac{72\sqrt{2}}{6} = 12\sqrt{2}
\]
### Final Answer
The final answer is:
\[
\boxed{12\sqrt{2}}
\]
