What is the foci ((x+1)^2}{64}+\frac{(y-5)^2)/9 =1 ?

The foci ((x+1)^2}{64}+\frac{(y-5)^2)/9 =1 is (-1+sqrt(55),5),(-1-sqrt(55),5)

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foci ((x+1)^2}{64}+\frac{(y-5)^2)/9 =1

foci

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 5 ) ^{2}}{9} = 1

(- 1 + 55, 5), (- 1 - 55, 5)

Solution steps

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

Calculate ellipse properties

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 5 ) ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (- 1, 5), a = 8, b = 3

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

Compute

c :

c = 8^{2} - 3^{2} : 55

(- 1 + 55, 5), (- 1 - 55, 5)

\frac{( x + 2 ) ^{2}}{16} + \frac{y ^{2}}{49} = 1

16 x^{2} + 192 x + 576 + 25 y^{2} = 1600

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

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

5 x^{2} + 9 y^{2} + 30 x = 0

What is the foci ((x+1)^2}{64}+\frac{(y-5)^2)/9 =1 ?
The foci ((x+1)^2}{64}+\frac{(y-5)^2)/9 =1 is (-1+sqrt(55),5),(-1-sqrt(55),5)