What is the foci ((x+1)^2)/(25)+((y+1)^2)/(49)=1 ?

The foci ((x+1)^2)/(25)+((y+1)^2)/(49)=1 is (-1,-1+2sqrt(6)),(-1,-1-2sqrt(6))

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foci ((x+1)^2)/(25)+((y+1)^2)/(49)=1

foci

\frac{( x + 1 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{49} = 1

(- 1, - 1 + 26), (- 1, - 1 - 26)

Solution steps

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

Calculate ellipse properties

\frac{( x + 1 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (- 1, - 1), b = 7, a = 5

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

Compute

c :

c = 7^{2} - 5^{2} : 26

(- 1, - 1 + 26), (- 1, - 1 - 26)

4 x^{2} + 36 y^{2} = 144

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

1 x^{2} + 4 y^{2} + 10 x + 9 y - 10 = 0

\frac{( x + 2 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{25} = 1

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 1 ) ^{2}}{100} = 1

What is the foci ((x+1)^2)/(25)+((y+1)^2)/(49)=1 ?
The foci ((x+1)^2)/(25)+((y+1)^2)/(49)=1 is (-1,-1+2sqrt(6)),(-1,-1-2sqrt(6))