What is the foci 4x^2-8x+y^2+4y-92=0 ?

The foci 4x^2-8x+y^2+4y-92=0 is (1,-2+5sqrt(3)),(1,-2-5sqrt(3))

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foci 4x^2-8x+y^2+4y-92=0

foci

4 x^{2} - 8 x + y^{2} + 4 y - 92 = 0

(1, - 2 + 53), (1, - 2 - 53)

Solution steps

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

Calculate ellipse properties

4 x^{2} - 8 x + y^{2} + 4 y - 92 = 0 :

Ellipse with center

(h, k) = (1, - 2), b = 10, a = 5

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

Compute

c :

c = 1 0^{2} - 5^{2} : 53

(1, - 2 + 53), (1, - 2 - 53)

\frac{x ^{2}}{9} + (y - 1)^{2} = 1

\frac{x ^{2}}{8} + \frac{y ^{2}}{24} = 1

\frac{( x - 2 ) ^{2}}{49} + \frac{( y + 1 ) ^{2}}{25} = 1

25 x^{2} + 9 y^{2} + 100 x - 125 = 0

4 x^{2} + 25 y^{2} - 40 x + 150 y + 225 = 0

What is the foci 4x^2-8x+y^2+4y-92=0 ?
The foci 4x^2-8x+y^2+4y-92=0 is (1,-2+5sqrt(3)),(1,-2-5sqrt(3))