What is the foci 25x^2+4y^2+50x-16y=59 ?

The foci 25x^2+4y^2+50x-16y=59 is (-1,2+sqrt(21)),(-1,2-sqrt(21))

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foci 25x^2+4y^2+50x-16y=59

foci

25 x^{2} + 4 y^{2} + 50 x - 16 y = 59

(- 1, 2 + 21), (- 1, 2 - 21)

Solution steps

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

Calculate ellipse properties

25 x^{2} + 4 y^{2} + 50 x - 16 y = 59 :

Ellipse with center

(h, k) = (- 1, 2), b = 5, a = 2

(- 1, 2 + c), (- 1, 2 - c)

Compute

c :

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

(- 1, 2 + 21), (- 1, 2 - 21)

3 x^{2} + 7 y^{2} - 21 = 0

9 x^{2} + 6 y^{2} + 36 y = 0

\frac{( x - 1 ) ^{2}}{12} + \frac{y ^{2}}{8} = 100

\frac{x ^{2}}{64} + \frac{y ^{2}}{81} = 1

x^{2} + y^{2} - 8 x - 20 y + 115 = 0

What is the foci 25x^2+4y^2+50x-16y=59 ?
The foci 25x^2+4y^2+50x-16y=59 is (-1,2+sqrt(21)),(-1,2-sqrt(21))