What is the foci ((x-5)^2)/4-((y+5)^2)/9 =1 ?

The foci ((x-5)^2)/4-((y+5)^2)/9 =1 is (5+sqrt(13),-5),(5-sqrt(13),-5)

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foci ((x-5)^2)/4-((y+5)^2)/9 =1

foci

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

(5 + 13, - 5), (5 - 13, - 5)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

(5 + c, - 5), (5 - c, - 5)

Compute

c :

c = 2^{2} + 3^{2} : 13

(5 + 13, - 5), (5 - 13, - 5)

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

\frac{y ^{2}}{16} - \frac{x ^{2}}{49} = 1

What is the foci ((x-5)^2)/4-((y+5)^2)/9 =1 ?
The foci ((x-5)^2)/4-((y+5)^2)/9 =1 is (5+sqrt(13),-5),(5-sqrt(13),-5)