What is the foci ((x-6)^2)/8-((y-5)^2)/1 =1 ?

The foci ((x-6)^2)/8-((y-5)^2)/1 =1 is (9,5),(3,5)

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foci ((x-6)^2)/8-((y-5)^2)/1 =1

foci

\frac{( x - 6 ) ^{2}}{8} - \frac{( y - 5 ) ^{2}}{1} = 1

(9, 5), (3, 5)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 6 ) ^{2}}{8} - \frac{( y - 5 ) ^{2}}{1} = 1 :

Right-left Hyperbola with

(h, k) = (6, 5), a = 22, b = 1

(6 + c, 5), (6 - c, 5)

Compute

c :

c = (22)^{2} + 1^{2} : 3

(6 + 3, 5), (6 - 3, 5)

Refine

(9, 5), (3, 5)

\frac{x ^{2}}{16} - \frac{( y + 5 ) ^{2}}{9} = 1

4 x^{2} = y^{2} + 4

x^{2} - 4 y^{2} - 2 x + 16 y = 20

\frac{x ^{2}}{45} - \frac{y ^{2}}{360} = 1

\frac{y ^{2}}{225} - \frac{x ^{2}}{64} = 1

What is the foci ((x-6)^2)/8-((y-5)^2)/1 =1 ?
The foci ((x-6)^2)/8-((y-5)^2)/1 =1 is (9,5),(3,5)