For the ellipse $25 x^{2} + 9 y^{2} - 150 x - 90 y + 225 = 0,$ the eccentricity $e$ is equal to:

\frac{2}{5}

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\frac{3}{5}

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\frac{4}{5}

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\frac{1}{5}

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Solution

The correct option is B $\frac{4}{5}$
$25 x^{2} + 9 y^{2} - 150 x - 90 y + 225 = 0$
$25 (x^{2} - 6 x) + 9 (y^{2} - 10 y) + 225 = 0$
$25 (x^{2} - 6 x + 9) + 9 (y^{2} - 10 y + 25) = 225$
$\Rightarrow \frac{(x - 3)^{2}}{9} + \frac{(y - 5)^{2}}{25} = 1$
Hence eccentricity $= \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$

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