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)

\frac{( x - 6 ) ^{2}}{36} + \frac{( y + 4 ) ^{2}}{16} = 1

1 = \frac{( x - 2 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{25}

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

\frac{( x + 2 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{16} = 1

12 (x + 4)^{2} + 3 (y - 1)^{2} = 48

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))