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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

x^{2} + 4 x + 4 y^{2} = 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 ^{2}}{\frac{16}{9}} + \frac{y ^{2}}{16} = 1

(\frac{x}{5})^{2} + (\frac{y}{3})^{2} = 1

9 x^{2} + 16 y^{2} = 1

4 (x - 4)^{2} + 9 y^{2} = 36

\frac{( x - 1 ) ^{2}}{45} + \frac{( y - 7 ) ^{2}}{81} = 1

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