Question

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

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{3}{7}$
• D. $\frac{4}{9}$

Hint:

For eccentricity, always divide the smaller squared semi-axis by the larger one inside the square root.

Answer 1

The Correct Option is B

Step 1: Concept
Complete the square to bring the equation into standard form $\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1$.

Step 2: Meaning
$9x^2 + 5(y^2 - 6y + 9) = 45 \implies 9x^2 + 5(y-3)^2 = 45$.
$\frac{x^2}{5} + \frac{(y-3)^2}{9} = 1$.

Step 3: Analysis
Here $a^2 = 9$ and $b^2 = 5$ (vertical ellipse since $a^2 > b^2$ under $y$).
$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{5}{9}}$.

Step 4: Conclusion
$e = \sqrt{4/9} = 2/3$. Final Answer: (B)