What is the foci ((x-5)^2}{16}-\frac{(y+7)^2)/9 =-1 ?

The foci ((x-5)^2}{16}-\frac{(y+7)^2)/9 =-1 is (5,-2),(5,-12)

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foci ((x-5)^2}{16}-\frac{(y+7)^2)/9 =-1

foci

\frac{( x - 5 ) ^{2}}{16} - \frac{( y + 7 ) ^{2}}{9} = - 1

(5, - 2), (5, - 12)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 5 ) ^{2}}{16} - \frac{( y + 7 ) ^{2}}{9} = - 1 :

Up-down Hyperbola with

(h, k) = (5, - 7), a = 3, b = 4

(5, - 7 + c), (5, - 7 - c)

Compute

c :

c = 3^{2} + 4^{2} : 5

(5, - 7 + 5), (5, - 7 - 5)

Refine

(5, - 2), (5, - 12)

18 x^{2} - 32 y^{2} - 108 y - 128 x - 254 = 0

\frac{y ^{2}}{15} - \frac{x ^{2}}{1} = 1

x^{2} - y^{2} + 7 x + 8 y = 5

- x^{2} + y^{2} = 1

9 (y - 2)^{2} - 4 x^{2} = 36

What is the foci ((x-5)^2}{16}-\frac{(y+7)^2)/9 =-1 ?
The foci ((x-5)^2}{16}-\frac{(y+7)^2)/9 =-1 is (5,-2),(5,-12)