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

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

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

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 4, 2), a = 3, b = 2

(- 4 + c, 2), (- 4 - c, 2)

Compute

c :

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

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

9 x^{2} + 49 y^{2} + 18 x + 294 y - 9 = 0

9 x^{2} + 7 y^{2} - 63 = 0

16 x^{2} + 64 y^{2} = 128

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

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

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