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

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

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

foci

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

(1 + 5, - 2), (1 - 5, - 2)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

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

(1 + 5, - 2), (1 - 5, - 2)

9 x^{2} - y^{2} - 54 x - 2 y + 71 = 0

25 x^{2} + 56476 x - 25 y^{2} - 3 y = 48838

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

10 x^{2} - 25 y^{2} = 100

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

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