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

The foci ((x+2)^2)/9+((y-1)^2)/(25)=1 is (-2,5),(-2,-3)

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 2, 1), b = 5, a = 3

(- 2, 1 + c), (- 2, 1 - c)

Compute

c :

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

(- 2, 1 + 4), (- 2, 1 - 4)

Refine

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

25 x^{2} + 4 y^{2} = 100

9 x^{2} + 36 y^{2} + 36 x + 432 y = - 1008

4 x^{2} + 9 y^{2} - 32 x - 36 y + 64 = 0

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

9 x^{2} - 54 x + y^{2} + 2 y + 46 = 0

What is the foci ((x+2)^2)/9+((y-1)^2)/(25)=1 ?
The foci ((x+2)^2)/9+((y-1)^2)/(25)=1 is (-2,5),(-2,-3)