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

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

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

foci

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

(- 4, - 1 + 23), (- 4, - 1 - 23)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 4, - 1), b = 4, a = 2

(- 4, - 1 + c), (- 4, - 1 - c)

Compute

c :

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

(- 4, - 1 + 23), (- 4, - 1 - 23)

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

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

\frac{x ^{2}}{5} + \frac{y ^{2}}{16} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{8} = 1

9 x^{2} + 4 y^{2} - 54 x + 40 y + 145 = 0

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