What is the foci ((x+4)^2)/4+((y+3)^2)/9 =1 ?

The foci ((x+4)^2)/4+((y+3)^2)/9 =1 is (-4,-3+sqrt(5)),(-4,-3-sqrt(5))

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foci ((x+4)^2)/4+((y+3)^2)/9 =1

foci

\frac{( x + 4 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1

(- 4, - 3 + 5), (- 4, - 3 - 5)

Solution steps

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

Calculate ellipse properties

\frac{( x + 4 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1 :

Ellipse with center

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

(- 4, - 3 + c), (- 4, - 3 - c)

Compute

c :

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

(- 4, - 3 + 5), (- 4, - 3 - 5)

2 x^{2} + 4 y^{2} + 8 x + 9 y - 3 = 0

36 x^{2} + 4 y^{2} + 360 x - 32 y + 820 = 0

\frac{( x - 1 ) ^{2}}{100} + \frac{( y - 2 ) ^{2}}{36} = 1

\frac{( x + 6 ) ^{2}}{4} + \frac{( y + 6 ) ^{2}}{25} = 1

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

What is the foci ((x+4)^2)/4+((y+3)^2)/9 =1 ?
The foci ((x+4)^2)/4+((y+3)^2)/9 =1 is (-4,-3+sqrt(5)),(-4,-3-sqrt(5))