The eccentricity of the ellipse `ax ^(2) + by ^(2) + 2 fx + 2gy + c =0` if axis of ellipse parallel to x -

Options

A. `sqrt (((b -a)/(b )))`
B. `sqrt (((a +b)/(b )))`
C. `sqrt (((a +b)/(a)))`
D. none of these

Answer

To find the eccentricity of the ellipse given by the equation \( ax^2 + by^2 + 2fx + 2gy + c = 0 \) when the axes of the ellipse are parallel to the x-axis, we can follow these steps:

### Step 1: Identify the standard form of the ellipse
The standard form of a horizontal ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \( a \) and \( b \) are the semi-major and semi-minor axes respectively.

### Step 2: Rewrite the given equation
We start with the equation:
\[ ax^2 + by^2 + 2fx + 2gy + c = 0 \]
To convert this into the standard form, we need to complete the square for the \( x \) and \( y \) terms.

### Step 3: Complete the square for \( x \)
For the \( x \) terms:
\[ ax^2 + 2fx = a\left(x^2 + \frac{2f}{a}x\right) \]
Now, we complete the square:
\[ x^2 + \frac{2f}{a}x = \left(x + \frac{f}{a}\right)^2 - \frac{f^2}{a^2} \]
Thus,
\[ ax^2 + 2fx = a\left(\left(x + \frac{f}{a}\right)^2 - \frac{f^2}{a^2}\right) = a\left(x + \frac{f}{a}\right)^2 - \frac{f^2}{a} \]

### Step 4: Complete the square for \( y \)
For the \( y \) terms:
\[ by^2 + 2gy = b\left(y^2 + \frac{2g}{b}y\right) \]
Completing the square gives:
\[ y^2 + \frac{2g}{b}y = \left(y + \frac{g}{b}\right)^2 - \frac{g^2}{b^2} \]
Thus,
\[ by^2 + 2gy = b\left(\left(y + \frac{g}{b}\right)^2 - \frac{g^2}{b^2}\right) = b\left(y + \frac{g}{b}\right)^2 - \frac{g^2}{b} \]

### Step 5: Substitute back into the equation
Substituting back into the original equation gives:
\[ a\left(x + \frac{f}{a}\right)^2 - \frac{f^2}{a} + b\left(y + \frac{g}{b}\right)^2 - \frac{g^2}{b} + c = 0 \]
Rearranging this, we have:
\[ a\left(x + \frac{f}{a}\right)^2 + b\left(y + \frac{g}{b}\right)^2 = \frac{f^2}{a} + \frac{g^2}{b} - c \]

### Step 6: Divide by the constant on the right
Let \( k = \frac{f^2}{a} + \frac{g^2}{b} - c \). Then we can write:
\[ \frac{a\left(x + \frac{f}{a}\right)^2}{k} + \frac{b\left(y + \frac{g}{b}\right)^2}{k} = 1 \]
This implies:
\[ \frac{\left(x + \frac{f}{a}\right)^2}{\frac{k}{a}} + \frac{\left(y + \frac{g}{b}\right)^2}{\frac{k}{b}} = 1 \]

### Step 7: Identify \( a^2 \) and \( b^2 \)
From the above, we can identify:
\[ a^2 = \frac{k}{a}, \quad b^2 = \frac{k}{b} \]

### Step 8: Calculate the eccentricity
The eccentricity \( e \) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values of \( a^2 \) and \( b^2 \):
\[ e = \sqrt{1 - \frac{\frac{k}{b}}{\frac{k}{a}}} = \sqrt{1 - \frac{a}{b}} \]

### Final Answer
Thus, the eccentricity of the ellipse is:
\[ e = \sqrt{1 - \frac{a}{b}} \]