This solution will guide you through each step of solving the problem, focusing on clarity, the underlying mathematical concepts, and common pitfalls.

---

Understanding the Problem and Key Concepts

The problem asks us to find the value of an expression involving the eccentricity and area of a specific triangle on an ellipse. This requires us to:

1. Identify the properties of the given ellipse (semi-axes, eccentricity, foci).
2. Find a specific point P on the ellipse based on a tangent condition.
3. Calculate the area of the triangle formed by P and the two foci.

We will use the following key concepts:

• Standard equation of an ellipse: ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$
• Eccentricity ($e$): For an ellipse with major axis along the x-axis, $b^2 = a^2(1 - e^2)$.
• Foci of an ellipse: $(\pm ae, 0)$ for major axis along x-axis.
• Equation of a tangent to an ellipse at $(x_1, y_1)$: ${{x{x_1}} \over {{a^2}}} + {{y{y_1}} \over {{b^2}}} = 1$
• Slope of a line: For $Ax + By + C = 0$, the slope is $-A/B$.
• Perpendicular lines: If two lines have slopes $m_1$ and $m_2$, they are perpendicular if $m_1 m_2 = -1$.
• Area of a triangle: For a triangle with vertices $P(x_1, y_1)$, $S(x_2, y_2)$, $S'(x_3, y_3)$, the area can be calculated as $A = {1 \over 2} \times \text{base} \times \text{height}$ if the base lies on an axis.

---

Step 1: Analyze the Ellipse and Determine its Fundamental Properties

The given equation of the ellipse is: ${{{x^2}} \over 8} + {{{y^2}} \over 4} = 1$

Explanation: Our first step is to compare this equation with the standard form of an ellipse centered at the origin, ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$. This allows us to identify the squares of the semi-major and semi-minor axes.

1. Identify $a^2$ and $b^2$: From the given equation, we have $a^2 = 8$ and $b^2 = 4$.

2. Determine the orientation of the major axis: Since $a^2 = 8 > b^2 = 4$, this means the semi-major axis is along the x-axis. This is crucial for correctly applying formulas for eccentricity and foci.

   • Tip: If $b^2 > a^2$, the major axis would be along the y-axis, and the formulas for eccentricity and foci would change accordingly ($a^2 = b^2(1-e^2)$ and foci $(0, \pm be)$).

3. Calculate semi-major and semi-minor axes:

   • Semi-major axis, $a = \sqrt{8} = 2\sqrt{2}$.
   • Semi-minor axis, $b = \sqrt{4} = 2$.

4. Calculate the eccentricity ($e$): Concept: For an ellipse with its major axis along the x-axis, the eccentricity $e$ is related to $a$ and $b$ by the formula $b^2 = a^2(1 - e^2)$. Why: Eccentricity is a measure of how "stretched" the ellipse is. It's essential for locating the foci. Substitute the values of $a^2$ and $b^2$: $4 = 8(1 - e^2)$ Divide by 8: ${1 \over 2} = 1 - e^2$ Rearrange to find $e^2$: $e^2 = 1 - {1 \over 2} = {1 \over 2}$ So, $e = {1 \over {\sqrt 2 }}$.

5. Determine the coordinates of the foci (S and S'): Concept: For an ellipse with the major axis along the x-axis, the foci are located at $(\pm ae, 0)$. Why: The foci are two fixed points used in the definition of an ellipse. We need their coordinates to calculate the area of triangle SPS'. Calculate the product $ae$: $ae = (2\sqrt{2}) \cdot \left( {1 \over {\sqrt 2 }} \right) = 2$ Therefore, the foci are $S(2, 0)$ and $S'(-2, 0)$.

---

Step 2: Find the Coordinates of Point P

We are given that point P is in the second quadrant and the tangent at P to the ellipse is perpendicular to the line $x + 2y = 0$.

1. Find the slope of the given line: The equation of the line is $x + 2y = 0$. Concept: To find the slope, we rewrite the equation in the slope-intercept form $y = mx + c$, where $m$ is the slope. $2y = -x$ $y = -{1 \over 2}x$ The slope of this line is $m_1 = -{1 \over 2}$.

2. Find the slope of the tangent at P: Concept: If two lines are perpendicular, the product of their slopes is $-1$. Why: We are given a perpendicularity condition, so we use this to find the slope of the tangent. Let the slope of the tangent at P be $m_t$. $m_t \cdot m_1 = -1$ $m_t \cdot \left( -{1 \over 2} \right) = -1$ $m_t = 2$

3. Find the coordinates of P using the tangent's slope: Let $P(x_1, y_1)$ be the point on the ellipse. Concept: The slope of the tangent to the ellipse ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ at a point $(x_1, y_1)$ is given by $m_t = -{{{b^2}x_1} \over {{a^2}y_1}}$. Why: This formula directly relates the slope of the tangent to the coordinates of the point of tangency, which will help us find $x_1$ and $y_1$. Substitute $a^2 = 8$ and $b^2 = 4$: $m_t = -{{4x_1} \over {8y_1}} = -{{x_1} \over {2y_1}}$ We know $m_t = 2$, so: $2 = -{{x_1} \over {2y_1}}$ Cross-multiply: $4y_1 = -x_1 \implies x_1 = -4y_1 \quad (*)$

4. Use the fact that P lies on the ellipse: Concept: Any point $(x_1, y_1)$ on the ellipse must satisfy its equation. Why: This gives us a second equation relating $x_1$ and $y_1$, allowing us to solve for their unique values. Substitute $P(x_1, y_1)$ into the ellipse equation: ${{x_1^2} \over 8} + {{y_1^2} \over 4} = 1$ Now, substitute $x_1 = -4y_1$ from equation $(*)$ into the ellipse equation: ${{(-4y_1)^2} \over 8} + {{y_1^2} \over 4} = 1$ ${{16y_1^2} \over 8} + {{y_1^2} \over 4} = 1$ $2y_1^2 + {{y_1^2} \over 4} = 1$ To combine terms, find a common denominator: ${{8y_1^2 + y_1^2} \over 4} = 1$ ${{9y_1^2} \over 4} = 1$ $9y_1^2 = 4$ $y_1^2 = {4 \over 9} \implies y_1 = \pm {2 \over 3}$

5. Determine the correct coordinates for P: Using $x_1 = -4y_1$:

   • If $y_1 = {2 \over 3}$, then $x_1 = -4 \left( {2 \over 3} \right) = -{8 \over 3}$. So, $P_1 \left( -{8 \over 3}, {2 \over 3} \right)$.
   • If $y_1 = -{2 \over 3}$, then $x_1 = -4 \left( -{2 \over 3} \right) = {8 \over 3}$. So, $P_2 \left( {8 \over 3}, -{2 \over 3} \right)$.

   Constraint: The problem states that P is in the second quadrant. Why: The second quadrant is where $x < 0$ and $y > 0$. Comparing $P_1$ and $P_2$ with this condition:

   • $P_1 \left( -{8 \over 3}, {2 \over 3} \right)$ has $x < 0$ and $y > 0$. This matches the second quadrant.
   • $P_2 \left( {8 \over 3}, -{2 \over 3} \right)$ has $x > 0$ and $y < 0$. This is in the fourth quadrant.

   Therefore, the coordinates of point P are $\left( -{8 \over 3}, {2 \over 3} \right)$.

---

Step 3: Calculate the Area of Triangle SPS'

We have the coordinates of the vertices:

• $P\left( -{8 \over 3}, {2 \over 3} \right)$
• $S(2, 0)$
• $S'(-2, 0)$

Concept: The foci $S$ and $S'$ lie on the x-axis, making the segment $S'S$ the base of the triangle. The length of this base is the distance between the foci, $2ae$. The height of the triangle with respect to this base is the perpendicular distance from P to the x-axis, which is simply $|y_1|$. Why: This method (base $\times$ height) is much simpler than the determinant formula when the base lies on an axis.

1. Calculate the base length: Base $S'S = 2ae = 2 \times 2 = 4$. Alternatively, distance between $(-2, 0)$ and $(2, 0)$ is $|2 - (-2)| = |4| = 4$.

2. Calculate the height: The height $h$ is the absolute value of the y-coordinate of P. $h = |y_P| = \left| {2 \over 3} \right| = {2 \over 3}$.

3. Calculate the area A: Concept: Area $A = {1 \over 2} \times \text{base} \times \text{height}$. $A = {1 \over 2} \times 4 \times {2 \over 3}$ $A = 2 \times {2 \over 3}$ $A = {4 \over 3}$

---

Step 4: Calculate the Final Expression

We need to find the value of $(5 - e^2) \cdot A$.

1. Recall the values: We found $e^2 = {1 \over 2}$ and $A = {4 \over 3}$.

2. Substitute and calculate: $(5 - e^2) \cdot A = \left( 5 - {1 \over 2} \right) \cdot {4 \over 3}$ $= \left( {{10 - 1} \over 2} \right) \cdot {4 \over 3}$ $= \left( {9 \over 2} \right) \cdot {4 \over 3}$ $= {9 \times 4 \over 2 \times 3}$ $= {{36} \over 6}$ $= 6$

---

Summary and Key Takeaway

The value of $(5 - e^2) \cdot A$ is $\boxed{6}$.

This problem effectively tests your understanding of various aspects of ellipses:

• Identifying properties from the standard equation.
• Calculating eccentricity and foci.
• Applying tangent conditions (slope, perpendicularity).
• Using coordinate