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

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

(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)

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

y^{2} - 4 x^{2} + 4 x - 2 y - 4 = 0

5 x^{2} - 4 y^{2} + 80 = 0

+ 9 x^{2} - y^{2} - 36 x - 6 y + 18 = 0

\frac{( y + 5 ) ^{2}}{16} - \frac{( x + 3 ) ^{2}}{9} = 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)