What is the foci ((x+1)^2)/4-((y-2)^2)/(12)=1 ?

The foci ((x+1)^2)/4-((y-2)^2)/(12)=1 is (3,2),(-5,2)

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foci ((x+1)^2)/4-((y-2)^2)/(12)=1

foci

\frac{( x + 1 ) ^{2}}{4} - \frac{( y - 2 ) ^{2}}{12} = 1

(3, 2), (- 5, 2)

Solution steps

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

Calculate Hyperbola properties

\frac{( x + 1 ) ^{2}}{4} - \frac{( y - 2 ) ^{2}}{12} = 1 :

Right-left Hyperbola with

(h, k) = (- 1, 2), a = 2, b = 23

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

Compute

c :

c = 2^{2} + (23)^{2} : 4

(- 1 + 4, 2), (- 1 - 4, 2)

Refine

(3, 2), (- 5, 2)

\frac{( y - 3 ) ^{2}}{7} - \frac{( x + 6 ) ^{2}}{42} = 1

\frac{x ^{2}}{3} - \frac{y ^{2}}{3} = 1

2 x^{2} - 3 y^{2} = 6

49 y^{2} - x^{2} = 49

\frac{x ^{2}}{81} - \frac{y ^{2}}{1 6 ^{2}} = 1

What is the foci ((x+1)^2)/4-((y-2)^2)/(12)=1 ?
The foci ((x+1)^2)/4-((y-2)^2)/(12)=1 is (3,2),(-5,2)