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

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

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

foci

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

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

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Compute

c :

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

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

- 25 x^{2} + 4 y^{2} + 150 x + 24 y - 289 = 0

9 x^{2} - 16 y^{2} + 36 x + 32 y - 124 = 0

- \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

100 x^{2} + 112952 x - 100 y^{2} - 106 y = 48838

4 x^{2} - 25 y^{2} + 16 x + 50 y - 109 = 0

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