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

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

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Compute

c :

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

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

\frac{x ^{2}}{64} - \frac{y ^{2}}{32} = 1

25 y^{2} - 81 x^{2} = 2025

\frac{y ^{2}}{4} - \frac{x ^{2}}{16} = 1

\frac{( y + 3 ) ^{2}}{64} - \frac{( x + 6 ) ^{2}}{49} = 1

\frac{( x + 4 ) ^{2}}{4} - \frac{( y - 1 ) ^{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,-1+sqrt(14)),(2,-1-sqrt(14))