Step 1: Understanding the Concept:
To find the eccentricity, we need to convert the general quadratic equation into the standard form of an ellipse: $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$. This is done by completing the square.
Step 2: Key Formula or Approach:
For an ellipse $\frac{X^2}{a^2} + \frac{Y^2}{b^2} = 1$:
If $a>b$ (horizontal major axis), $e = \sqrt{1 - \frac{b^2}{a^2}}$.
If $b>a$ (vertical major axis), $e = \sqrt{1 - \frac{a^2}{b^2}}$.
Step 3: Detailed Explanation:
Given equation:
\[ 9x^2 + 5y^2 - 30y = 0 \]
Group terms with the same variable and factor out the leading coefficient for the $y$ terms:
\[ 9x^2 + 5(y^2 - 6y) = 0 \]
Complete the square inside the parenthesis. Take half of the $y$ coefficient (-6), square it, and add/subtract it: $(-6/2)^2 = (-3)^2 = 9$.
\[ 9x^2 + 5(y^2 - 6y + 9 - 9) = 0 \]
\[ 9x^2 + 5((y - 3)^2 - 9) = 0 \]
Distribute the 5:
\[ 9x^2 + 5(y - 3)^2 - 45 = 0 \]
\[ 9x^2 + 5(y - 3)^2 = 45 \]
Divide the entire equation by 45 to make the right side 1:
\[ \frac{9x^2}{45} + \frac{5(y - 3)^2}{45} = 1 \]
\[ \frac{x^2}{5} + \frac{(y - 3)^2}{9} = 1 \]
This is the standard form of an ellipse. Comparing with $\frac{x^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$:
Here, the denominator of the $x^2$ term is 5, and the denominator of the $y^2$ term is 9.
Since $9>5$, the major axis is vertical.
Let $a^2 = 5$ (semi-minor axis squared) and $b^2 = 9$ (semi-major axis squared).
The formula for eccentricity when the major axis is vertical is $e^2 = 1 - \frac{a^2}{b^2}$:
\[ e^2 = 1 - \frac{5}{9} \]
\[ e^2 = \frac{9 - 5}{9} = \frac{4}{9} \]
Taking the positive square root:
\[ e = \frac{2}{3} \]
Step 4: Final Answer:
The eccentricity is $\frac{2}{3}$.