What is the foci ((x+2)^2)/(16)+((y+2)^2)/(25)=1 ?

The foci ((x+2)^2)/(16)+((y+2)^2)/(25)=1 is (-2,1),(-2,-5)

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foci ((x+2)^2)/(16)+((y+2)^2)/(25)=1

foci

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

(- 2, 1), (- 2, - 5)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 2, - 2), b = 5, a = 4

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

Compute

c :

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

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

Refine

(- 2, 1), (- 2, - 5)

\frac{x ^{2}}{96} + \frac{y ^{2}}{100} = 1

9 x^{2} + 25 y^{2} - 100 y - 125 = 0

3 x^{2} + 4 y^{2} = 12

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

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

What is the foci ((x+2)^2)/(16)+((y+2)^2)/(25)=1 ?
The foci ((x+2)^2)/(16)+((y+2)^2)/(25)=1 is (-2,1),(-2,-5)