consider a branch of the hyperbola x^2-2y^2-2√2x-4√2y-6=0 with vertex at the point A.Let B be one of the end points of its lactus rectum.Let C be the focus of the hyperbola nearest to the point A then 1)the area of triangle abc 2)2e^2-3=0 3)center of the hperbola 4)Length of LR

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.

Consider a branch of the hyperbola x^2-2y^2-2√2x-4√2y-6=0 with vertex at the point A. Let B be one of the endpoints of its latus rectum. Let C be the focus of the hyperbola nearest to the point A. Then: 1) Find the area of triangle ABC, 2) Solve for 2e^2 - 3 = 0, 3) Find the center of the hyperbola, 4) Find the length of the latus rectum.

Learn how to solve this complex hyperbola problem by identifying the center, calculating the length of the latus rectum, and finding the area of triangle ABC formed by a vertex, a focus, and a point on the latus rectum. Suitable for advanced students in Grades 11-12.

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