An ellipse passing through `(4sqrt2, 2 sqrt 6)` foci at (-4,0) and (4,0). Its eccentricity is

Options

A. `sqrt(2)`
B. `1/2`
C. `(1)/sqrt(2)`
D. `(1)/sqrt(3)`

Answer

To find the eccentricity of the ellipse that passes through the point \((4\sqrt{2}, 2\sqrt{6})\) and has foci at \((-4, 0)\) and \((4, 0)\), we can follow these steps:

### Step 1: Identify the values of \(a\) and the center of the ellipse
The foci of the ellipse are given as \((-4, 0)\) and \((4, 0)\). The center of the ellipse is the midpoint of the foci, which is at the origin \((0, 0)\). The distance from the center to each focus is \(c = 4\).

### Step 2: Use the relationship between \(a\), \(b\), and \(c\)
For an ellipse, we have the relationship:
\[ c^2 = a^2 - b^2 \]
where \(c\) is the distance from the center to the foci, \(a\) is the semi-major axis, and \(b\) is the semi-minor axis.

### Step 3: Set up the equation of the ellipse
The standard form of the ellipse with a horizontal major axis is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
We will use the point \((4\sqrt{2}, 2\sqrt{6})\) to find \(a\) and \(b\).

### Step 4: Substitute the point into the ellipse equation
Substituting the point into the ellipse equation gives:
\[ \frac{(4\sqrt{2})^2}{a^2} + \frac{(2\sqrt{6})^2}{b^2} = 1 \]
Calculating the squares:
\[ \frac{32}{a^2} + \frac{24}{b^2} = 1 \]

### Step 5: Express \(b^2\) in terms of \(a^2\)
From the relationship \(c^2 = a^2 - b^2\):
\[ 16 = a^2 - b^2 \implies b^2 = a^2 - 16 \]

### Step 6: Substitute \(b^2\) into the ellipse equation
Substituting \(b^2\) into the equation:
\[ \frac{32}{a^2} + \frac{24}{a^2 - 16} = 1 \]

### Step 7: Clear the denominators
Multiply through by \(a^2(a^2 - 16)\):
\[ 32(a^2 - 16) + 24a^2 = a^2(a^2 - 16) \]
This simplifies to:
\[ 32a^2 - 512 + 24a^2 = a^4 - 16a^2 \]
Combining like terms:
\[ a^4 - 72a^2 + 512 = 0 \]

### Step 8: Let \(t = a^2\) and solve the quadratic equation
We rewrite the equation:
\[ t^2 - 72t + 512 = 0 \]
Using the quadratic formula:
\[ t = \frac{72 \pm \sqrt{72^2 - 4 \cdot 512}}{2} \]
Calculating the discriminant:
\[ 72^2 - 2048 = 5184 - 2048 = 3136 \]
Taking the square root:
\[ \sqrt{3136} = 56 \]
Thus:
\[ t = \frac{72 \pm 56}{2} \]
Calculating the two possible values:
\[ t_1 = \frac{128}{2} = 64 \quad \text{and} \quad t_2 = \frac{16}{2} = 8 \]

### Step 9: Determine \(a\) from \(t\)
Thus, we have:
\[ a^2 = 64 \quad \text{or} \quad a^2 = 8 \]
Calculating \(a\):
\[ a = 8 \quad \text{or} \quad a = 2\sqrt{2} \]

### Step 10: Calculate eccentricity \(e\)
Using \(c = 4\):
1. If \(a = 8\):
\[ e = \frac{c}{a} = \frac{4}{8} = \frac{1}{2} \]
2. If \(a = 2\sqrt{2}\):
\[ e = \frac{c}{a} = \frac{4}{2\sqrt{2}} = \sqrt{2} \quad (\text{not valid since } e \geq 1) \]

### Conclusion
Thus, the eccentricity of the ellipse is:
\[ \boxed{\frac{1}{2}} \]