What is the foci-((y-3)^2}{25}+\frac{(x+2)^2)/9 =1 ?

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

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

foci

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

(- 2 + 34, 3), (- 2 - 34, 3)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

(- 2 + c, 3), (- 2 - c, 3)

Compute

c :

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

(- 2 + 34, 3), (- 2 - 34, 3)

\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

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

\frac{x ^{2}}{81} - \frac{y ^{2}}{9} = 1

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