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

The vertices ((x-2)^2)/4-((y-2)^2)/5 =1 is (4,2),(0,2)

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vertices ((x-2)^2)/4-((y-2)^2)/5 =1

vertices

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

(4, 2), (0, 2)

Solution steps

(h + a, k), (h - a, k)

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

(2 + 2, 2), (2 - 2, 2)

Refine

(4, 2), (0, 2)

\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

8 x^{2} - 2 y^{2} = 16

What is the vertices ((x-2)^2)/4-((y-2)^2)/5 =1 ?
The vertices ((x-2)^2)/4-((y-2)^2)/5 =1 is (4,2),(0,2)