What is the foci ((y+5)^2)/(64)-((x+2)^2)/(36)=1 ?

The foci ((y+5)^2)/(64)-((x+2)^2)/(36)=1 is (-2,5),(-2,-15)

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foci ((y+5)^2)/(64)-((x+2)^2)/(36)=1

foci

\frac{( y + 5 ) ^{2}}{64} - \frac{( x + 2 ) ^{2}}{36} = 1

(- 2, 5), (- 2, - 15)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 5 ) ^{2}}{64} - \frac{( x + 2 ) ^{2}}{36} = 1 :

Up-down Hyperbola with

(h, k) = (- 2, - 5), a = 8, b = 6

(- 2, - 5 + c), (- 2, - 5 - c)

Compute

c :

c = 8^{2} + 6^{2} : 10

(- 2, - 5 + 10), (- 2, - 5 - 10)

Refine

(- 2, 5), (- 2, - 15)

\frac{x ^{2}}{1} - \frac{y ^{2}}{35} = 1

x^{2} - 36 y^{2} = 36

9 x^{2} - 9 y^{2} + 18 x + 36 y = 63

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

64 y^{2} - 4 x^{2} = 256

What is the foci ((y+5)^2)/(64)-((x+2)^2)/(36)=1 ?
The foci ((y+5)^2)/(64)-((x+2)^2)/(36)=1 is (-2,5),(-2,-15)