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

The foci ((x+5)^2)/4-(y^2)/4 =1 is (-5+2sqrt(2),0),(-5-2sqrt(2),0)

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

foci

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

(- 5 + 22, 0), (- 5 - 22, 0)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (- 5, 0), a = 2, b = 2

(- 5 + c, 0), (- 5 - c, 0)

Compute

c :

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

(- 5 + 22, 0), (- 5 - 22, 0)

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

- 25 x^{2} + 4 y^{2} + 150 x + 24 y - 289 = 0

9 x^{2} - 16 y^{2} + 36 x + 32 y - 124 = 0

- \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

100 x^{2} + 112952 x - 100 y^{2} - 106 y = 48838

What is the foci ((x+5)^2)/4-(y^2)/4 =1 ?
The foci ((x+5)^2)/4-(y^2)/4 =1 is (-5+2sqrt(2),0),(-5-2sqrt(2),0)