What is the foci ((x+2)^2)/4+((y-3)^2)/9 =1 ?

The foci ((x+2)^2)/4+((y-3)^2)/9 =1 is (-2,3+sqrt(5)),(-2,3-sqrt(5))

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foci ((x+2)^2)/4+((y-3)^2)/9 =1

foci

\frac{( x + 2 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{9} = 1

(- 2, 3 + 5), (- 2, 3 - 5)

Solution steps

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

Calculate ellipse properties

\frac{( x + 2 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (- 2, 3), b = 3, a = 2

(- 2, 3 + c), (- 2, 3 - c)

Compute

c :

c = 3^{2} - 2^{2} : 5

(- 2, 3 + 5), (- 2, 3 - 5)

4 x^{2} + 8 y^{2} = 3

16 x^{2} + 4 y^{2} - 64 = 0

\frac{( x + 6 ) ^{2}}{7} + \frac{( y - 3 ) ^{2}}{16} = 1

\frac{( x - 4 ) ^{2}}{3} + \frac{y ^{2}}{9} = 1

\frac{x ^{2}}{36} + \frac{8 y ^{2}}{81} = 1

What is the foci ((x+2)^2)/4+((y-3)^2)/9 =1 ?
The foci ((x+2)^2)/4+((y-3)^2)/9 =1 is (-2,3+sqrt(5)),(-2,3-sqrt(5))