To solve this problem, let’s proceed step-by-step:

Step 1: Rewrite the Hyperbola Equation

The equation of the hyperbola is given as: $x^2 - 2y^2 - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0$ Let's rearrange terms and complete the square.

1. Complete the square for $x$-terms and $y$-terms:

Rewrite the equation as: $x^2 - 2\sqrt{2}x - 2y^2 - 4\sqrt{2}y = 6$

For $x$-terms: $x^2 - 2\sqrt{2}x = (x - \sqrt{2})^2 - 2$

For $y$-terms: $-2y^2 - 4\sqrt{2}y = -2\left(y^2 + 2\sqrt{2}y\right) = -2\left((y + \sqrt{2})^2 - 2\right) = -2(y + \sqrt{2})^2 + 4$

Substitute back into the equation: $(x - \sqrt{2})^2 - 2(y + \sqrt{2})^2 = 6 + 2 - 4 = 4$

Divide by 4: $\frac{(x - \sqrt{2})^2}{4} - \frac{(y + \sqrt{2})^2}{2} = 1$

This is the equation of a hyperbola with center at $(\sqrt{2}, -\sqrt{2})$.

Step 2: Identify Parameters of the Hyperbola

The hyperbola equation is now in standard form: $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$ where $(h, k) = (\sqrt{2}, -\sqrt{2})$, $a^2 = 4$, and $b^2 = 2$.

• Center: $(\sqrt{2}, -\sqrt{2})$
• Vertices: $a = 2$, so vertices are at $(\sqrt{2} \pm 2, -\sqrt{2})$.
• Foci: The distance of each focus from the center is $c$, where $c^2 = a^2 + b^2 = 4 + 2 = 6$. Thus, $c = \sqrt{6}$.

The foci are located at: $(\sqrt{2} \pm \sqrt{6}, -\sqrt{2})$

Step 3: Find the Length of the Latus Rectum (LR)

The length of the latus rectum of a hyperbola is given by: $\text{Length of LR} = \frac{2b^2}{a} = \frac{2 \cdot 2}{2} = 2$

Step 4: Area of Triangle ABC

Let's label the points:

• A (Vertex): Choose one vertex, say $(\sqrt{2} + 2, -\sqrt{2})$.
• B (End of Latus Rectum): Since the latus rectum passes through the focus, take $B = (\sqrt{2} + \sqrt{6}, -\sqrt{2} + 1)$ as one end of the latus rectum.
• C (Focus): $C = (\sqrt{2} + \sqrt{6}, -\sqrt{2})$.

Using the coordinates of $A$, $B$, and $C$, apply the formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$: $\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$

Substitute the coordinates:

• $A = (\sqrt{2} + 2, -\sqrt{2})$
• $B = (\sqrt{2} + \sqrt{6}, -\sqrt{2} + 1)$
• $C = (\sqrt{2} + \sqrt{6}, -\sqrt{2})$

The area calculation simplifies further.

Step 5: Solve for $2e^2 - 3$

The eccentricity $e$ of a hyperbola is given by $e = \frac{c}{a}$: $e = \frac{\sqrt{6}}{2} = \frac{\sqrt{6}}{2}$

Thus: $2e^2 = 2 \left(\frac{\sqrt{6}}{2}\right)^2 = 2 \cdot \frac{6}{4} = 3$ So, $2e^2 - 3 = 0$, which holds true.