An ellipse has foci (4, 2), (2, 2) and it passes through the point P (2, 4). The eccentricity of the ellipse is

Options

A. `tan.(pi)/(10)`
B. `tan.(pi)/(12)`
C. `tan.(pi)/(6)`
D. `tan.(pi)/(8)`

Answer

To find the eccentricity of the ellipse given the foci and a point through which it passes, we can follow these steps:

### Step 1: Identify the foci and the point
The foci of the ellipse are given as \( S_1(4, 2) \) and \( S_2(2, 2) \). The point through which the ellipse passes is \( P(2, 4) \).

### Step 2: Calculate the distance between the foci
The distance \( d \) between the foci \( S_1 \) and \( S_2 \) can be calculated using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of the foci:
\[ d = \sqrt{(2 - 4)^2 + (2 - 2)^2} = \sqrt{(-2)^2 + 0^2} = \sqrt{4} = 2 \]

### Step 3: Relate the distance to the semi-major axis and eccentricity
The distance between the foci is given by \( 2c \), where \( c \) is the distance from the center of the ellipse to each focus. Therefore:
\[ 2c = 2 \implies c = 1 \]
The relationship between \( a \) (semi-major axis), \( b \) (semi-minor axis), and \( c \) is given by:
\[ c^2 = a^2 - b^2 \]

### Step 4: Use the point on the ellipse to find \( a \)
The ellipse passes through the point \( P(2, 4) \). The condition for an ellipse states that the sum of the distances from any point on the ellipse to the two foci is constant and equal to \( 2a \). We need to calculate \( S_1P \) and \( S_2P \).

Calculating \( S_1P \):
\[ S_1P = \sqrt{(2 - 4)^2 + (4 - 2)^2} = \sqrt{(-2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

Calculating \( S_2P \):
\[ S_2P = \sqrt{(2 - 2)^2 + (4 - 2)^2} = \sqrt{0^2 + (2)^2} = \sqrt{4} = 2 \]

Now, we can find \( S_1P + S_2P \):
\[ S_1P + S_2P = 2\sqrt{2} + 2 \]

### Step 5: Set the sum equal to \( 2a \)
From the property of the ellipse:
\[ S_1P + S_2P = 2a \implies 2\sqrt{2} + 2 = 2a \]
Dividing by 2:
\[ \sqrt{2} + 1 = a \]

### Step 6: Calculate \( b \) using \( c \) and \( a \)
We know \( c = 1 \) and \( a = \sqrt{2} + 1 \). We can now find \( b \):
\[ c^2 = a^2 - b^2 \implies 1^2 = (\sqrt{2} + 1)^2 - b^2 \]
Calculating \( a^2 \):
\[ (\sqrt{2} + 1)^2 = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2} \]
Thus:
\[ 1 = (3 + 2\sqrt{2}) - b^2 \implies b^2 = 3 + 2\sqrt{2} - 1 = 2 + 2\sqrt{2} \]

### Step 7: Calculate the eccentricity \( e \)
The eccentricity \( e \) is given by:
\[ e = \frac{c}{a} \]
Substituting the values:
\[ e = \frac{1}{\sqrt{2} + 1} \]

### Step 8: Rationalizing the denominator
To express \( e \) in a more standard form:
\[ e = \frac{1}{\sqrt{2} + 1} \cdot \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{1} = \sqrt{2} - 1 \]

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