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

The vertices ((x-6)^2)/8-((y-5)^2)/1 =1 is (6+2sqrt(2),5),(6-2sqrt(2),5)

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

vertices

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

(6 + 22, 5), (6 - 22, 5)

Solution steps

(h + a, k), (h - a, 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 + 22, 5), (6 - 22, 5)

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

\frac{x ^{2} - y ^{2}}{24} = 1

x^{2} + 6 x = 9 y^{2}

\frac{y ^{2}}{4 8 ^{2}} - \frac{x ^{2}}{2 0 ^{2}} = 1

x^{2} - y^{2} + 2 x + 2 y - 16 = 0

What is the vertices ((x-6)^2)/8-((y-5)^2)/1 =1 ?
The vertices ((x-6)^2)/8-((y-5)^2)/1 =1 is (6+2sqrt(2),5),(6-2sqrt(2),5)