What is the foci ((x+3)^2)/4+((y+2)^2)/(16)=1 ?

The foci ((x+3)^2)/4+((y+2)^2)/(16)=1 is (-3,-2+2sqrt(3)),(-3,-2-2sqrt(3))

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foci ((x+3)^2)/4+((y+2)^2)/(16)=1

foci

\frac{( x + 3 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{16} = 1

(- 3, - 2 + 23), (- 3, - 2 - 23)

Solution steps

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

Calculate ellipse properties

\frac{( x + 3 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{16} = 1 :

Ellipse with center

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

(- 3, - 2 + c), (- 3, - 2 - c)

Compute

c :

c = 4^{2} - 2^{2} : 23

(- 3, - 2 + 23), (- 3, - 2 - 23)

1 x^{(2)} + 4 y^{(2)} + 10 x + 9 y - 10 = 0

64 x^{2} + 16 y^{2} = 1024

16 x^{2} + 36 y^{2} + 192 x - 288 y + 576 = 0

4 x^{2} - 8 x + y^{2} + 4 y - 92 = 0

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

What is the foci ((x+3)^2)/4+((y+2)^2)/(16)=1 ?
The foci ((x+3)^2)/4+((y+2)^2)/(16)=1 is (-3,-2+2sqrt(3)),(-3,-2-2sqrt(3))