What is the foci ((y+4)^2)/9-((x+2)^2)/4 =1 ?

The foci ((y+4)^2)/9-((x+2)^2)/4 =1 is (-2,-4+sqrt(13)),(-2,-4-sqrt(13))

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foci ((y+4)^2)/9-((x+2)^2)/4 =1

foci

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

(- 2, - 4 + 13), (- 2, - 4 - 13)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

(- 2, - 4 + c), (- 2, - 4 - c)

Compute

c :

c = 3^{2} + 2^{2} : 13

(- 2, - 4 + 13), (- 2, - 4 - 13)

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

\frac{y ^{2}}{25} - x^{2} = 1

\frac{x ^{2}}{16} - \frac{y ^{2}}{48} = 1

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

\frac{( y + 3 ) ^{2}}{9} - \frac{( x - 1 ) ^{2}}{18} = 1

What is the foci ((y+4)^2)/9-((x+2)^2)/4 =1 ?
The foci ((y+4)^2)/9-((x+2)^2)/4 =1 is (-2,-4+sqrt(13)),(-2,-4-sqrt(13))