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

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

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

foci

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

(1, - 2 + 3), (1, - 2 - 3)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

c = 2^{2} - 1^{2} : 3

(1, - 2 + 3), (1, - 2 - 3)

\frac{x ^{2}}{16} + \frac{y ^{2}}{64} = 1

81 x^{2} + 4 y^{2} = 324

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 5 ) ^{2}}{9} = 1

\frac{( x + 2 ) ^{2}}{16} + \frac{y ^{2}}{49} = 1

16 x^{2} + 192 x + 576 + 25 y^{2} = 1600

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