What is the foci ((x+1)^2)/(28)+((y+5)^2)/(23)=1 ?

The foci ((x+1)^2)/(28)+((y+5)^2)/(23)=1 is (-1+sqrt(5),-5),(-1-sqrt(5),-5)

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foci ((x+1)^2)/(28)+((y+5)^2)/(23)=1

foci

\frac{( x + 1 ) ^{2}}{28} + \frac{( y + 5 ) ^{2}}{23} = 1

(- 1 + 5, - 5), (- 1 - 5, - 5)

Solution steps

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

Calculate ellipse properties

\frac{( x + 1 ) ^{2}}{28} + \frac{( y + 5 ) ^{2}}{23} = 1 :

Ellipse with center

(h, k) = (- 1, - 5), a = 27, b = 23

(- 1 + c, - 5), (- 1 - c, - 5)

Compute

c :

c = (27)^{2} - 23^{2} : 5

(- 1 + 5, - 5), (- 1 - 5, - 5)

x^{2} - 8 x + 25 y^{2} - 100 y + 91 = 0

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

2 x^{2} + y^{2} = 50

\frac{( x - 5 ) ^{2}}{8 ^{2}} + \frac{( y + 2 ) ^{2}}{1 0 ^{2}} = 1

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

What is the foci ((x+1)^2)/(28)+((y+5)^2)/(23)=1 ?
The foci ((x+1)^2)/(28)+((y+5)^2)/(23)=1 is (-1+sqrt(5),-5),(-1-sqrt(5),-5)