What is the foci ((y+5)^2}{16}-\frac{(x+3)^2)/9 =1 ?

The foci ((y+5)^2}{16}-\frac{(x+3)^2)/9 =1 is (-3,0),(-3,-10)

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foci ((y+5)^2}{16}-\frac{(x+3)^2)/9 =1

foci

\frac{( y + 5 ) ^{2}}{16} - \frac{( x + 3 ) ^{2}}{9} = 1

(- 3, 0), (- 3, - 10)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 5 ) ^{2}}{16} - \frac{( x + 3 ) ^{2}}{9} = 1 :

Up-down Hyperbola with

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

(- 3, - 5 + c), (- 3, - 5 - c)

Compute

c :

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

(- 3, - 5 + 5), (- 3, - 5 - 5)

Refine

(- 3, 0), (- 3, - 10)

\frac{( y + 5 ) ^{2}}{\frac{1}{9}} - \frac{( x - 3 ) ^{2}}{\frac{1}{4}} = 1

x^{2} - y^{2} - 4 x + 4 y + 9 = 0

\frac{( x + 3 ) ^{2}}{25} - \frac{( y - 2 ) ^{2}}{49} = 1

4 y^{2} - x^{2} = 900

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

What is the foci ((y+5)^2}{16}-\frac{(x+3)^2)/9 =1 ?
The foci ((y+5)^2}{16}-\frac{(x+3)^2)/9 =1 is (-3,0),(-3,-10)