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

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

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

foci

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

(- 1 + 22, - 2), (- 1 - 22, - 2)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

c = 2^{2} + 2^{2} : 22

(- 1 + 22, - 2), (- 1 - 22, - 2)

x^{2} - y^{2} - 10 x - 10 y - 1 = 0

\frac{( x - 3 ) ^{2}}{16} - \frac{( y + 8 ) ^{2}}{64} = 1

9 x^{2} - 4 y^{2} - 18 x - 16 y + 29 = 0

9 y^{2} + 36 y - 4 x^{2} - 24 x - 36 = 0

4 (x - 3)^{2} - 9 (y - 2)^{2} = 36

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