Step 1: Convert the equation of the hyperbola to standard form.

The given equation is ${x^2} - 2{y^2} - 2 \sqrt 2 x - 4\sqrt 2 y - 6 = 0$. To convert this to the standard form, we complete the square for the x and y terms.

Group the terms: $(x^2 - 2\sqrt{2}x) - 2(y^2 + 2\sqrt{2}y) - 6 = 0$

Complete the square for x: $(x^2 - 2\sqrt{2}x + (\sqrt{2})^2 - (\sqrt{2})^2) = (x - \sqrt{2})^2 - 2$

Complete the square for y: $-2(y^2 + 2\sqrt{2}y + (\sqrt{2})^2 - (\sqrt{2})^2) = -2((y + \sqrt{2})^2 - 2) = -2(y + \sqrt{2})^2 + 4$

Substitute these back into the equation: $((x - \sqrt{2})^2 - 2) + (-2(y + \sqrt{2})^2 + 4) - 6 = 0$ $(x - \sqrt{2})^2 - 2(y + \sqrt{2})^2 - 4 = 0$ $(x - \sqrt{2})^2 - 2(y + \sqrt{2})^2 = 4$

Divide by 4 to get the standard form $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$: $\frac{(x - \sqrt{2})^2}{4} - \frac{(y + \sqrt{2})^2}{2} = 1$

Step 2: Identify the parameters of the hyperbola.

From the standard form, we can identify:

• Center $(h, k) = (\sqrt{2}, -\sqrt{2})$
• $a^2 = 4 \implies a = 2$
• $b^2 = 2 \implies b = \sqrt{2}$

Now, we calculate the eccentricity, $e$: $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{2}{4}} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}}$

Step 3: Determine the coordinates of the points A, B, and C.

• Point A (Vertex): The vertices of the hyperbola are at $(h \pm a, k)$. We are considering one branch. Let's take the right branch. The vertex $A$ is: $A = (h+a, k) = (\sqrt{2} + 2, -\sqrt{2})$

• Point C (Focus nearest to A): The foci are at $(h \pm ae, k)$. First, calculate $ae$: $ae = 2 \cdot \sqrt{\frac{3}{2}} = \sqrt{4 \cdot \frac{3}{2}} = \sqrt{6}$. The foci are $S_1 = (\sqrt{2} + \sqrt{6}, -\sqrt{2})$ and $S_2 = (\sqrt{2} - \sqrt{6}, -\sqrt{2})$. The vertex A is on the right branch, so the nearest focus will also be on the right side of the center. Thus, C is $S_1$. $C = (\sqrt{2} + \sqrt{6}, -\sqrt{2})$

• Point B (Endpoint of Latus Rectum): The latus rectum is a line segment passing through a focus and perpendicular to the transverse axis. Its endpoints are $(h \pm ae, k \pm \frac{b^2}{a})$. Since C is the focus, the endpoints of the latus rectum through C have coordinates $(x_C, y_C \pm \frac{b^2}{a})$. Calculate the semi-latus rectum length: $\frac{b^2}{a} = \frac{2}{2} = 1$. The endpoints are $(\sqrt{2} + \sqrt{6}, -\sqrt{2} \pm 1)$. Let's choose one of these points as B: $B = (\sqrt{2} + \sqrt{6}, -\sqrt{2} + 1)$

Step 4: Calculate the area of triangle ABC.

The coordinates of the vertices of the triangle are: $A = (\sqrt{2} + 2, -\sqrt{2})$ $B = (\sqrt{2} + \sqrt{6}, -\sqrt{2} + 1)$ $C = (\sqrt{2} + \sqrt{6}, -\sqrt{2})$

Notice that points A and C have the same y-coordinate, so the side AC is horizontal. Points B and C have the same x-coordinate, so the side BC is vertical. Therefore, triangle ABC is a right-angled triangle with the right angle at C.

Calculate the lengths of the base AC and the height BC:

• Length of base AC: $AC = |x_C - x_A| = |(\sqrt{2} + \sqrt{6}) - (\sqrt{2} + 2)| = |\sqrt{6} - 2| = \sqrt{6} - 2 \quad (\text{since } \sqrt{6} > 2)$
• Length of height BC: $BC = |y_B - y_C| = |(-\sqrt{2} + 1) - (-\sqrt{2})| = |1| = 1$

Now, calculate the area of the triangle: $Area(\triangle ABC) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times BC$ $Area = \frac{1}{2} \times (\sqrt{6} - 2) \times 1 = \frac{\sqrt{6} - 2}{2} = \frac{\sqrt{6}}{2} - 1$

To match the options, we can rewrite $\frac{\sqrt{6}}{2}$ as $\sqrt{\frac{6}{4}} = \sqrt{\frac{3}{2}}$. So, the area is $\sqrt{\frac{3}{2}} - 1$.

This matches option B.