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

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Solution

The correct option is D
6

$25 x^{2} + 16 y^{2} + 100 x - 64 y - 236$ = 0

$25 (x^{2} + 4 x) + 16 (y^{2} - 4 y)$ = 236

$25 (x + 2)^{2} + 16 (y - 2)^{2}$ = 400

$\frac{(x + 2)^{2}}{16} + \frac{(y + 2)^{2}}{25}$ = 1, e = $\sqrt{1 - \frac{16}{25}}$ = $\frac{3}{5}$

$S S^{1}$ = $2 b e$ = $2 \times 5 \times (\frac{3}{5})$ = 6

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25 x^{2} + 9 y^{2} - 150 x - 90 y + 225 = 0

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(4 \sqrt{3}, 5)

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