What is the foci ((x+1)^2)/(100)+((y+2)^2)/(121)=1 ?

The foci ((x+1)^2)/(100)+((y+2)^2)/(121)=1 is (-1,-2+sqrt(21)),(-1,-2-sqrt(21))

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foci ((x+1)^2)/(100)+((y+2)^2)/(121)=1

foci

\frac{( x + 1 ) ^{2}}{100} + \frac{( y + 2 ) ^{2}}{121} = 1

(- 1, - 2 + 21), (- 1, - 2 - 21)

Solution steps

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

Calculate ellipse properties

\frac{( x + 1 ) ^{2}}{100} + \frac{( y + 2 ) ^{2}}{121} = 1 :

Ellipse with center

(h, k) = (- 1, - 2), b = 11, a = 10

(- 1, - 2 + c), (- 1, - 2 - c)

Compute

c :

c = 1 1^{2} - 1 0^{2} : 21

(- 1, - 2 + 21), (- 1, - 2 - 21)

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{4} = 1

\frac{( x - 4 ) ^{2}}{24} + \frac{( y - 2 ) ^{2}}{49} = 1

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

\frac{( x + 13 ) ^{2}}{400} + \frac{( y + 6 ) ^{2}}{625} = 1

4 x^{2} + y^{2} - 16 x - 2 y - 19 = 0

What is the foci ((x+1)^2)/(100)+((y+2)^2)/(121)=1 ?
The foci ((x+1)^2)/(100)+((y+2)^2)/(121)=1 is (-1,-2+sqrt(21)),(-1,-2-sqrt(21))