What is the foci ((y+4)^2)/9-((x-3)^2)/(25)=1 ?

The foci ((y+4)^2)/9-((x-3)^2)/(25)=1 is (3,-4+sqrt(34)),(3,-4-sqrt(34))

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foci ((y+4)^2)/9-((x-3)^2)/(25)=1

foci

\frac{( y + 4 ) ^{2}}{9} - \frac{( x - 3 ) ^{2}}{25} = 1

(3, - 4 + 34), (3, - 4 - 34)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 4 ) ^{2}}{9} - \frac{( x - 3 ) ^{2}}{25} = 1 :

Up-down Hyperbola with

(h, k) = (3, - 4), a = 3, b = 5

(3, - 4 + c), (3, - 4 - c)

Compute

c :

c = 3^{2} + 5^{2} : 34

(3, - 4 + 34), (3, - 4 - 34)

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

- \frac{( y - 3 ) ^{2}}{25} + \frac{( x + 2 ) ^{2}}{9} = 1

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 5 ) ^{2}}{9} = 1

2 x^{2} - 2 y^{2} - 20 x + 16 y + 16 = 0

25 y^{2} - 16 x^{2} - 400 = 0

What is the foci ((y+4)^2)/9-((x-3)^2)/(25)=1 ?
The foci ((y+4)^2)/9-((x-3)^2)/(25)=1 is (3,-4+sqrt(34)),(3,-4-sqrt(34))