The vertices of a quadrilateral are situated at foci and the extrimities of the minor axis of the ellipse `4x^(2) + 9y^(2) = 36 .` Find the area of the quadrilateral .

Answer

To find the area of the quadrilateral formed by the foci and the extremities of the minor axis of the ellipse given by the equation \(4x^2 + 9y^2 = 36\), we will follow these steps:

### Step 1: Convert the equation of the ellipse to standard form
The given equation is:
\[ 4x^2 + 9y^2 = 36 \]
Dividing the entire equation by 36, we get:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
This shows that \(a^2 = 9\) and \(b^2 = 4\).

### Step 2: Identify the values of \(a\) and \(b\)
From \(a^2 = 9\) and \(b^2 = 4\):
\[ a = \sqrt{9} = 3, \quad b = \sqrt{4} = 2 \]

### Step 3: Calculate the distance to the foci
The distance \(c\) to the foci is given by the formula:
\[ c^2 = a^2 - b^2 \]
Substituting the values:
\[ c^2 = 9 - 4 = 5 \quad \Rightarrow \quad c = \sqrt{5} \]

### Step 4: Determine the coordinates of the foci and the extremities of the minor axis
The foci of the ellipse are located at \((\pm c, 0)\), which gives us:
\[ F_1 = (-\sqrt{5}, 0), \quad F_2 = (\sqrt{5}, 0) \]
The extremities of the minor axis are located at \((0, \pm b)\), which gives us:
\[ A = (0, 2), \quad B = (0, -2) \]

### Step 5: Identify the vertices of the quadrilateral
The vertices of the quadrilateral are:
\[ A(0, 2), \quad B(0, -2), \quad F_1(-\sqrt{5}, 0), \quad F_2(\sqrt{5}, 0) \]

### Step 6: Calculate the area of the quadrilateral
To find the area of the quadrilateral \(AF_1BF_2\), we can divide it into two triangles: \(AF_1B\) and \(BF_2A\).

#### Area of triangle \(AF_1B\):
Using the formula for the area of a triangle with vertices at \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For triangle \(AF_1B\):
- \(A(0, 2)\)
- \(F_1(-\sqrt{5}, 0)\)
- \(B(0, -2)\)

Substituting the coordinates:
\[ \text{Area}_{AF_1B} = \frac{1}{2} \left| 0(0 + 2) + (-\sqrt{5})(-2 - 2) + 0(2 - 0) \right| \]
\[ = \frac{1}{2} \left| 0 + 4\sqrt{5} + 0 \right| = \frac{1}{2} \cdot 4\sqrt{5} = 2\sqrt{5} \]

#### Area of triangle \(BF_2A\):
For triangle \(BF_2A\):
- \(B(0, -2)\)
- \(F_2(\sqrt{5}, 0)\)
- \(A(0, 2)\)

Substituting the coordinates:
\[ \text{Area}_{BF_2A} = \frac{1}{2} \left| 0(0 - 2) + \sqrt{5}(2 + 2) + 0(-2 - 0) \right| \]
\[ = \frac{1}{2} \left| 0 + 4\sqrt{5} + 0 \right| = \frac{1}{2} \cdot 4\sqrt{5} = 2\sqrt{5} \]

### Step 7: Total area of the quadrilateral
The total area of quadrilateral \(AF_1BF_2\) is:
\[ \text{Area}_{quadrilateral} = \text{Area}_{AF_1B} + \text{Area}_{BF_2A} = 2\sqrt{5} + 2\sqrt{5} = 4\sqrt{5} \]

### Final Answer
The area of the quadrilateral is:
\[ \boxed{4\sqrt{5}} \]