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

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

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

foci

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

(- 2, 4 + 23), (- 2, 4 - 23)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

(- 2, 4 + 23), (- 2, 4 - 23)

4 x^{2} + 25 y^{2} = 100

\frac{y ^{2}}{25} + \frac{x ^{2}}{9} = 1

4 x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

9 x^{2} - 36 x + 4 y^{2} = 0

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

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