Foci of the Ellipse 25x2 + 9y2 – 150 x – 90y + 225 = 0 are

25x-32+ 9y-52=225x-329+y-5225=10, ± 5 × 45= x-3, y-5= 3,1 and3,9

Foci of the Ellipse 25x2 + 9y2 – 150 x – 90y + 225 = 0 are

25x-32+ 9y-52=225x-329+y-5225=10, ± 5 × 45= x-3, y-5= 3,1 and3,9

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Foci of the Ellipse 25x²+ 9y² – 150 x – 90y + 225 = 0 are

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(5, 7), (4, 7)

(6, 9), (5, 1)

(4, 9), (4, 1)

(3, 9), (3, 1)

answer is A.

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Detailed Solution

$25 {(x-3)}^{2} + 9 {(y-5)}^{2} =225$
$\frac{{(x-3)}^{2}}{9} + \frac{{(y-5)}^{2}}{25} =1$
$(0, ± 5 × \frac{4}{5}) = (x-3, y-5) = (3,1) and (3,9)$

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Foci of the Ellipse 25x2 + 9y2 – 150 x – 90y + 225 = 0 are

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Foci of the Ellipse 25x²+ 9y² – 150 x – 90y + 225 = 0 are