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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

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

9 x^{2} + 25 y^{2} + 108 x - 200 y + 499 = 0

\frac{( x - 6 ) ^{2}}{49} + \frac{( y - 6 ) ^{2}}{36} = 1

8 x^{2} + 4 y^{2} + 16 x - 12 y - 15 = 0

\frac{25 x ^{2}}{4} - 50 x + 100 y^{2} + 800 y + 1075 = 0

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