Find the eccentricity and the coordinates of foci of the hyperbola $25 x^{2} - 9 y^{2} = 225$

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Solution

$25 x^{2} - 9 y^{2} = 225$
The given equation can be written as
$\frac{x^{2}}{9} - \frac{y^{2}}{25} = 1 . . . . (1)$
Here, $a^{2} = 9 \Rightarrow a = 3$ and $b^{2} = 25 \Rightarrow b = 5$
eccentricity, $e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{25}{9}}$
$= \sqrt{\frac{34}{9}} = \sqrt{\frac{34}{3}}$
$F o c i = (\pm a e, 0)$
$= (\pm 3 \times \frac{\sqrt{34}}{3}, 0) = (\pm \sqrt{34}, 0)$

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Find the eccentricity and the coordinates of foci of the hyperbola 25x2−9y2=225

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