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

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

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

foci

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

(- 2 + 3, 0), (- 2 - 3, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 2, 0), a = 2, b = 1

(- 2 + c, 0), (- 2 - c, 0)

Compute

c :

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

(- 2 + 3, 0), (- 2 - 3, 0)

\frac{( x - 1 ) ^{2}}{6} + \frac{( y + 2 ) ^{2}}{18} = 1

\frac{( x + 11 ) ^{2}}{100} + \frac{( y - 4 ) ^{2}}{1 9 ^{2}} = 1

\frac{( x + 6 ) ^{2}}{18} + \frac{( y + 3 ) ^{2}}{9} = 1

y^{2} + x^{2} = 100

\frac{x ^{2}}{k} + \frac{y ^{2}}{4 + k} = 1

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