What is the foci 25x^2+4y^2-250x-16y+541=0 ?

The foci 25x^2+4y^2-250x-16y+541=0 is (5,2+sqrt(21)),(5,2-sqrt(21))

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foci 25x^2+4y^2-250x-16y+541=0

foci

25 x^{2} + 4 y^{2} - 250 x - 16 y + 541 = 0

(5, 2 + 21), (5, 2 - 21)

Solution steps

(h, k + c), (h, k - c)

Calculate ellipse properties

25 x^{2} + 4 y^{2} - 250 x - 16 y + 541 = 0 :

Ellipse with center

(h, k) = (5, 2), b = 5, a = 2

(5, 2 + c), (5, 2 - c)

Compute

c :

c = 5^{2} - 2^{2} : 21

(5, 2 + 21), (5, 2 - 21)

\frac{x ^{2}}{25} + \frac{y ^{2}}{21} = 1

\frac{( x + 2 ) ^{2}}{2 ^{2}} + \frac{( y - 3 ) ^{2}}{3 ^{2}} = 1

81 x^{2} + 49 y^{2} = 3969

25 x^{2} + 4 y^{2} + 100 x = 0

\frac{x ^{2}}{9 ^{2}} + \frac{y ^{2}}{6 ^{2}} = 1

What is the foci 25x^2+4y^2-250x-16y+541=0 ?
The foci 25x^2+4y^2-250x-16y+541=0 is (5,2+sqrt(21)),(5,2-sqrt(21))