The eccentricity of the ellipse `9x^2+5y^2-30 y=0` is

Options

A. `1//3`
B. `2//3`
C. `3//4`
D. none of these

Answer

To find the eccentricity of the ellipse given by the equation \( 9x^2 + 5y^2 - 30y = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
Start with the equation:
\[ 9x^2 + 5y^2 - 30y = 0 \]
Rearranging gives:
\[ 9x^2 + 5y^2 = 30y \]

### Step 2: Complete the square for the \( y \) terms
We can rewrite the \( y \) terms by completing the square. First, factor out the 5 from the \( y \) terms:
\[ 9x^2 + 5(y^2 - 6y) = 0 \]
Now, complete the square for \( y^2 - 6y \):
\[ y^2 - 6y = (y - 3)^2 - 9 \]
Substituting back, we have:
\[ 9x^2 + 5((y - 3)^2 - 9) = 0 \]
This simplifies to:
\[ 9x^2 + 5(y - 3)^2 - 45 = 0 \]
Rearranging gives:
\[ 9x^2 + 5(y - 3)^2 = 45 \]

### Step 3: Divide by 45 to get the standard form
Now, divide the entire equation by 45:
\[ \frac{9x^2}{45} + \frac{5(y - 3)^2}{45} = 1 \]
This simplifies to:
\[ \frac{x^2}{5} + \frac{(y - 3)^2}{9} = 1 \]

### Step 4: Identify \( a^2 \) and \( b^2 \)
From the standard form of the ellipse:
\[ \frac{x^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
we can identify:
- \( a^2 = 5 \) (thus \( a = \sqrt{5} \))
- \( b^2 = 9 \) (thus \( b = 3 \))

### Step 5: Calculate the eccentricity
For an ellipse where \( b > a \), the eccentricity \( e \) is given by the formula:
\[ e = \sqrt{1 - \frac{a^2}{b^2}} \]
Substituting the values of \( a^2 \) and \( b^2 \):
\[ e = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{9 - 5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3} \]

### Final Answer
The eccentricity of the ellipse is:
\[ \frac{2}{3} \]