What is the foci ((x-5)/8)^2-((y+2)/7)^2=1 ?

The foci ((x-5)/8)^2-((y+2)/7)^2=1 is (5+sqrt(113),-2),(5-sqrt(113),-2)

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foci ((x-5)/8)^2-((y+2)/7)^2=1

foci

(\frac{x - 5}{8})^{2} - (\frac{y + 2}{7})^{2} = 1

(5 + 113, - 2), (5 - 113, - 2)

Solution steps

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

Calculate Hyperbola properties

(\frac{x - 5}{8})^{2} - (\frac{y + 2}{7})^{2} = 1 :

Right-left Hyperbola with

(h, k) = (5, - 2), a = 8, b = 7

(5 + c, - 2), (5 - c, - 2)

Compute

c :

c = 8^{2} + 7^{2} : 113

(5 + 113, - 2), (5 - 113, - 2)

3 x^{2} - y^{2} = 3

(\frac{x ^{2}}{16}) - (\frac{y ^{2}}{9}) = 1

\frac{x ^{2}}{34} - \frac{y ^{2}}{2} = 1

16 y^{2} - 49 x^{2} = 784

\frac{( y + 2 ) ^{2}}{25} - \frac{( x - 9 ) ^{2}}{81} = 1

What is the foci ((x-5)/8)^2-((y+2)/7)^2=1 ?
The foci ((x-5)/8)^2-((y+2)/7)^2=1 is (5+sqrt(113),-2),(5-sqrt(113),-2)