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

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

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foci ((x+5)^2)/9+((y-1)^2)/4 =1

foci

\frac{( x + 5 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{4} = 1

(- 5 + 5, 1), (- 5 - 5, 1)

Solution steps

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

Calculate ellipse properties

\frac{( x + 5 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{4} = 1 :

Ellipse with center

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

(- 5 + c, 1), (- 5 - c, 1)

Compute

c :

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

(- 5 + 5, 1), (- 5 - 5, 1)

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{3} = 1

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

16 x^{2} + 25 y^{2} - 128 x + 150 y + 81 = 0

x^{2} + 4 y^{2} + 8 y = 0

16 x^{2} + 25 y^{2} - 96 x - 256 = 0

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