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

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

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

foci

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

(1 + 5, - 3), (1 - 5, - 3)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

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

(1 + 5, - 3), (1 - 5, - 3)

144 y^{2} - 25 x^{2} = 3600

\frac{y ^{2}}{25} - \frac{x ^{2}}{39} = 1

16 y^{2} - 25 x^{2} = 1600

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

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

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