Distance between the foci of the ellipse $25 x^{2} + 16 y^{2} + 100 x - 64 y - 236 = 0$ is

8

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12

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9

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6

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Solution

The correct option is D $6$
$25 (x + 2)^{2} + 16 (y - 2)^{2} = 400$
$\frac{(x + 2)^{2}}{16} + \frac{(y - 2)^{2}}{25} = 1$
Distance between foci $= 2 b e$
$= 2 \times 5 \times \sqrt{\frac{25 - 16}{25}}$
$= 2 \times 3$
$= 6$

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