The eccentricity of the ellipse 9x^2 + 5y^2 - | vedclass

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(B) Given equation: $9x^2 + 5y^2 - 30y = 0$ Complete the square for $y$: $9x^2 + 5(y^2 - 6y) = 0$ $9x^2 + 5(y^2 - 6y + 9) = 45$ $9x^2 + 5(y - 3)^2 = 4

Vedclass · Accepted Answer

$2/3$

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(B) Given equation: $9x^2 + 5y^2 - 30y = 0$ Complete the square for $y$: $9x^2 + 5(y^2 - 6y) = 0$ $9x^2 + 5(y^2 - 6y + 9) = 45$ $9x^2 + 5(y - 3)^2 = 45$ Divide by $45$: $\frac{x^2}{5} + \frac{(y - 3)^2}{9} = 1$ Here,$a^2 = 5$ and $b^2 = 9$. Since $b^2 > a^2$,the ellipse is vertical. The eccentricity $e$ is given by $a^2 = b^2(1 - e^2)$ $5 = 9(1 - e^2)$ $1 - e^2 = \frac{5}{9}$ $e^2 = 1 - \frac{5}{9} = \frac{4}{9}$ $e = \sqrt{\frac{4}{9}} = \frac{2}{3}$

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

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