What is the foci ((x-2)^2)/5-((y+1)^2)/9 =1 ?

The foci ((x-2)^2)/5-((y+1)^2)/9 =1 is (2+sqrt(14),-1),(2-sqrt(14),-1)

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foci ((x-2)^2)/5-((y+1)^2)/9 =1

foci

(2 + 14, - 1), (2 - 14, - 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (2, - 1), a = 5, b = 3

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

Compute

c :

c = 5^{2} + 3^{2} : 14

(2 + 14, - 1), (2 - 14, - 1)

What is the foci ((x-2)^2)/5-((y+1)^2)/9 =1 ?
The foci ((x-2)^2)/5-((y+1)^2)/9 =1 is (2+sqrt(14),-1),(2-sqrt(14),-1)