Question

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

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. $\frac{4}{5}$

Answer 1

The correct option is B

$\frac{2}{3}$

Explanation of correct answer

Finding the eccentricity of the ellipse

Given, equation of the ellipse is $9{\mathrm{x}}^2+5{\mathrm{y}}^2-30\mathrm{y}=0$

Solving the equation,

$$\begin{gathered} \Rightarrow 9{\mathrm{x}}^2+5{\mathrm{y}}^2-30\mathrm{y}=0 \\ \Rightarrow 9{\mathrm{x}}^2+5{\mathrm{y}}^2-30\mathrm{y}+45=0+45 \\ \Rightarrow 9{\mathrm{x}}^2+5{\mathrm{y}}^2-30\mathrm{y}+45-45=0 \\ \Rightarrow 9{\mathrm{x}}^2+5\left[{\mathrm{y}}^2-6\mathrm{y}+9-9\right]=0 \\ \Rightarrow 9{\mathrm{x}}^2+5{\left[\mathrm{y}-3\right]}^2=45 \\ \Rightarrow \frac{9{\mathrm{x}}^2}{45}+\frac{5{\left[\mathrm{y}-3\right]}^2}{45}=\frac{45}{45} \\ \Rightarrow \frac{{\mathrm{x}}^2}{5}+\frac{{\left[\mathrm{y}-3\right]}^2}{9}=1\left[\mathrm{form}\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\right] \end{gathered}$$

As we know, eccentricity is given by,

$e=\sqrt{\frac{a^2-b^2}{a^2}}$

So,

$$\begin{gathered} e=\sqrt{1-\frac{b^2}{a^2}} \\ =\sqrt{1-\frac{5}{9}}\left(a^2=9,b^2=5\right) \\ =\sqrt{\frac{4}{9}} \\ =\frac{2}{3} \end{gathered}$$

Thus, the eccentricity of the ellipse is $\frac{2}{3}$.

Hence, the correct answer is Option (B).