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

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

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

foci

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

(5, - 5), (- 5, - 5)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, - 5), a = 4, b = 3

(0 + c, - 5), (0 - c, - 5)

Compute

c :

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

(0 + 5, - 5), (0 - 5, - 5)

Refine

(5, - 5), (- 5, - 5)

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

x^{2} - 4 y^{2} - 2 x + 16 y = 20

\frac{x ^{2}}{45} - \frac{y ^{2}}{360} = 1

\frac{y ^{2}}{225} - \frac{x ^{2}}{64} = 1

4 y^{2} - 9 x^{2} + 16 y + 18 x = 29

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