What is the foci ((y+3)^2)/4-((x+1)^2)/(16)=1 ?

The foci ((y+3)^2)/4-((x+1)^2)/(16)=1 is (-1,-3+2sqrt(5)),(-1,-3-2sqrt(5))

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foci ((y+3)^2)/4-((x+1)^2)/(16)=1

foci

\frac{( y + 3 ) ^{2}}{4} - \frac{( x + 1 ) ^{2}}{16} = 1

(- 1, - 3 + 25), (- 1, - 3 - 25)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 3 ) ^{2}}{4} - \frac{( x + 1 ) ^{2}}{16} = 1 :

Up-down Hyperbola with

(h, k) = (- 1, - 3), a = 2, b = 4

(- 1, - 3 + c), (- 1, - 3 - c)

Compute

c :

c = 2^{2} + 4^{2} : 25

(- 1, - 3 + 25), (- 1, - 3 - 25)

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

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

- 9 x^{2} + y^{2} + 18 x - 45 = 0

\frac{( y - 4 ) ^{2}}{25} - \frac{( x + 1 ) ^{2}}{4} = 1

+ \frac{x ^{2}}{49} - \frac{y ^{2}}{9} = 1

What is the foci ((y+3)^2)/4-((x+1)^2)/(16)=1 ?
The foci ((y+3)^2)/4-((x+1)^2)/(16)=1 is (-1,-3+2sqrt(5)),(-1,-3-2sqrt(5))