What is the foci ((x+2)^2)/9+((y+3)^2)/(25)=1 ?

The foci ((x+2)^2)/9+((y+3)^2)/(25)=1 is (-2,1),(-2,-7)

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foci ((x+2)^2)/9+((y+3)^2)/(25)=1

foci

\frac{( x + 2 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{25} = 1

(- 2, 1), (- 2, - 7)

Solution steps

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

Calculate ellipse properties

\frac{( x + 2 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (- 2, - 3), b = 5, a = 3

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

Compute

c :

c = 5^{2} - 3^{2} : 4

(- 2, - 3 + 4), (- 2, - 3 - 4)

Refine

(- 2, 1), (- 2, - 7)

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 1 ) ^{2}}{100} = 1

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

18 x^{2} + 24 x + 12 y^{2} - 12 y = 1

x^{2} + 25 y^{2} = 25

x^{2} + 9 y^{2} + 6 x - 90 y + 225 = 0

What is the foci ((x+2)^2)/9+((y+3)^2)/(25)=1 ?
The foci ((x+2)^2)/9+((y+3)^2)/(25)=1 is (-2,1),(-2,-7)