What is the foci (x+1)^2+(y^2)/4 =1 ?

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

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foci (x+1)^2+(y^2)/4 =1

foci

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

(- 1, 3), (- 1, - 3)

Solution steps

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

Calculate ellipse properties

(x + 1)^{2} + \frac{y ^{2}}{4} = 1 :

Ellipse with center

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

(- 1, 0 + c), (- 1, 0 - c)

Compute

c :

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

(- 1, 0 + 3), (- 1, 0 - 3)

Refine

(- 1, 3), (- 1, - 3)

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

\frac{( X - 1 ) ^{2}}{4} + \frac{( Y + 2 ) ^{2}}{16} = 1

25 x^{2} + 9 y^{2} - 54 y - 144 = 0

\frac{( x + 3 ) ^{2}}{49} + \frac{( y - 1 ) ^{2}}{48} = 1

36 x^{2} + 9 y^{2} = 324

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