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

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

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foci ((x+2)^2)/4-((y+1)^2)/5 =1

foci

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

(1, - 1), (- 5, - 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (- 2, - 1), a = 2, b = 5

(- 2 + c, - 1), (- 2 - c, - 1)

Compute

c :

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

(- 2 + 3, - 1), (- 2 - 3, - 1)

Refine

(1, - 1), (- 5, - 1)

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

x^{2} - 2 x - 3 = y^{2} + 2 y + 3

6 x^{2} - 5 y^{2} - 24 x + 10 y - 11 = 0

4 x^{2} - 24 x - 25 y^{2} + 250 y - 489 = 0

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

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