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

\frac{1}{3}

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\frac{2}{3}

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\frac{3}{4}

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None of these

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Solution

The correct option is D $\frac{2}{3}$
$9 x^{2} + 5 (y^{2} - 6 y) = 0$
$9 x^{2} + 5 (y^{2} - 6 y + 9) = 45$
$9 x^{2} + 5 (y - 3)^{2} = 45$
$\frac{x^{2}}{5} + \frac{(y - 3)^{2}}{9} = 1$
$a^{2} < b^{2}$
$a^{2} = b^{2} (1 - e^{2})$
$5 = 9 (1 - e^{2})$
$\frac{5}{9} = 1 - e^{2}$
$e^{2} = \frac{4}{9}$
$e = \frac{2}{3}$

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Similar questions

9 x^{2} + 4 y^{2} - 30 y = 0

\frac{1}{p} \sqrt{q}

p

q .

x^{2} + 3 y^{2} = a^{2}

5 x^{2} + 4 y^{2} = 1

9 x^{2} + 5 y^{2} - 30 y = 0

9 x^{2} + 5 y^{2} - 30 y = 0

9 x^{2} + 5 y^{2} = 30 y

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