The eccentricity of ellipse `ax^2 + by^2 + 2gx + 2fy + c = 0` if its axis is parallel to x-axis is

Options

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

Answer

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

### Step 1: Rearranging the Equation
Start with the given equation of the ellipse:
\[ ax^2 + by^2 + 2gx + 2fy + c = 0 \]
Rearranging it gives:
\[ ax^2 + 2gx + by^2 + 2fy + c = 0 \]

### Step 2: Completing the Square
We will complete the square for the \(x\) and \(y\) terms separately.

For the \(x\) terms:
\[ ax^2 + 2gx = a\left(x^2 + \frac{2g}{a}x\right) \]
Completing the square:
\[ = a\left(x + \frac{g}{a}\right)^2 - \frac{g^2}{a} \]

For the \(y\) terms:
\[ by^2 + 2fy = b\left(y^2 + \frac{2f}{b}y\right) \]
Completing the square:
\[ = b\left(y + \frac{f}{b}\right)^2 - \frac{f^2}{b} \]

### Step 3: Substitute Back
Substituting these completed squares back into the equation gives:
\[ a\left(x + \frac{g}{a}\right)^2 - \frac{g^2}{a} + b\left(y + \frac{f}{b}\right)^2 - \frac{f^2}{b} + c = 0 \]
Rearranging this leads to:
\[ a\left(x + \frac{g}{a}\right)^2 + b\left(y + \frac{f}{b}\right)^2 = \frac{g^2}{a} + \frac{f^2}{b} - c \]

### Step 4: Standard Form of the Ellipse
To express this in standard form, we divide through by the right-hand side:
\[ \frac{a\left(x + \frac{g}{a}\right)^2}{\frac{g^2}{a} + \frac{f^2}{b} - c} + \frac{b\left(y + \frac{f}{b}\right)^2}{\frac{g^2}{a} + \frac{f^2}{b} - c} = 1 \]

Let:
\[ A = \frac{g^2}{a} + \frac{f^2}{b} - c \]
Then we have:
\[ \frac{\left(x + \frac{g}{a}\right)^2}{\frac{A}{a}} + \frac{\left(y + \frac{f}{b}\right)^2}{\frac{A}{b}} = 1 \]

### Step 5: Identify \(a\) and \(b\)
From the standard form, we can identify:
\[ \text{Semi-major axis} = \sqrt{\frac{A}{a}}, \quad \text{Semi-minor axis} = \sqrt{\frac{A}{b}} \]

### Step 6: Calculate Eccentricity
The eccentricity \(e\) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{b}{a}} \]
Substituting \(a\) and \(b\) from our previous steps:
\[ e = \sqrt{1 - \frac{\frac{g^2}{a} + \frac{f^2}{b} - c}{\frac{g^2}{a} + \frac{f^2}{b} - c}} \]
This simplifies to:
\[ e = \sqrt{1 - \frac{b}{a}} \]

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