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

The foci ((y+5)^2)/9-((x+2)^2)/(16)=1 is (-2,0),(-2,-10)

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foci ((y+5)^2)/9-((x+2)^2)/(16)=1

foci

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

(- 2, 0), (- 2, - 10)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (- 2, - 5), a = 3, b = 4

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

Compute

c :

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

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

Refine

(- 2, 0), (- 2, - 10)

- x^{2} + \frac{y ^{2}}{20} = 1

x^{2} - 16 y^{2} = 144

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

\frac{y ^{2}}{18} - \frac{x ^{2}}{2} = 1

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

What is the foci ((y+5)^2)/9-((x+2)^2)/(16)=1 ?
The foci ((y+5)^2)/9-((x+2)^2)/(16)=1 is (-2,0),(-2,-10)