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

The foci (y+4)^2-((x-2)^2)/4 =1 is (2,-4+sqrt(5)),(2,-4-sqrt(5))

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

foci

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

(2, - 4 + 5), (2, - 4 - 5)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (2, - 4), a = 1, b = 2

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

Compute

c :

c = 1^{2} + 2^{2} : 5

(2, - 4 + 5), (2, - 4 - 5)

\frac{( x - 6 ) ^{2}}{8} - \frac{( y - 5 ) ^{2}}{1} = 1

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

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

What is the foci (y+4)^2-((x-2)^2)/4 =1 ?
The foci (y+4)^2-((x-2)^2)/4 =1 is (2,-4+sqrt(5)),(2,-4-sqrt(5))