What is the foci ((x+2)^2}{25}+\frac{(y-5)^2)/9 =1 ?

The foci ((x+2)^2}{25}+\frac{(y-5)^2)/9 =1 is (2,5),(-6,5)

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foci ((x+2)^2}{25}+\frac{(y-5)^2)/9 =1

foci

\frac{( x + 2 ) ^{2}}{25} + \frac{( y - 5 ) ^{2}}{9} = 1

(2, 5), (- 6, 5)

Solution steps

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

Calculate ellipse properties

\frac{( x + 2 ) ^{2}}{25} + \frac{( y - 5 ) ^{2}}{9} = 1 :

Ellipse with center

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

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

Compute

c :

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

(- 2 + 4, 5), (- 2 - 4, 5)

Refine

(2, 5), (- 6, 5)

\frac{( x + 4 ) ^{2}}{24} + \frac{( y + 2 ) ^{2}}{49} = 1

3 x^{2} + 4 y^{2} = 48

9 (x + 2)^{2} + 4 (y - 3)^{2} = 36

\frac{( x + 1 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{169} = 1

16 (x - 1)^{2} + 25 (y - 4)^{2} = 400

What is the foci ((x+2)^2}{25}+\frac{(y-5)^2)/9 =1 ?
The foci ((x+2)^2}{25}+\frac{(y-5)^2)/9 =1 is (2,5),(-6,5)