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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

5 x^{2} + 9 y^{2} + 30 x = 0

\frac{x ^{2}}{64} + \frac{y ^{2}}{9} = 1

\frac{x ^{2}}{20} + \frac{y ^{2}}{36} = 1

\frac{x ^{2}}{12} + \frac{y ^{2}}{6} = 1

\frac{( x + 4 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{16} = 1

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