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

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

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

foci

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

(5, - 1), (- 15, - 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (- 5, - 1), a = 6, b = 8

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

Compute

c :

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

(- 5 + 10, - 1), (- 5 - 10, - 1)

Refine

(5, - 1), (- 15, - 1)

x^{2} - y^{2} - 4 x - 2 y - 1 = 0

9 x^{2} - y^{2} - 54 x + 45 = 0

x^{2} - 16 y^{2} = 16

3 x^{2} - y^{2} = 11

\frac{y ^{2}}{1 2 ^{2}} - \frac{x ^{2}}{5 ^{2}} = 1

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