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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

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 - 9 ) ^{2}}{4} + \frac{( y + 10 ) ^{2}}{81} = 1

18 x^{2} + 8 y^{2} + 36 x - 32 y - 22 = 0

\frac{x ^{2}}{32} + \frac{y ^{2}}{9} = 1

4 x^{2} + 25 y^{2} - 24 x + 250 y + 561 = 0

14 x^{2} - 70 = - 5 y^{2}

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