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

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

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

foci

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

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

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

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

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

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

\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

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