What is the vertices ((y-4)/4)^2-((x-6)/4)^2=1 ?

The vertices ((y-4)/4)^2-((x-6)/4)^2=1 is (6,8),(6,0)

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vertices ((y-4)/4)^2-((x-6)/4)^2=1

vertices

(\frac{y - 4}{4})^{2} - (\frac{x - 6}{4})^{2} = 1

(6, 8), (6, 0)

Solution steps

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

Calculate Hyperbola properties

(\frac{y - 4}{4})^{2} - (\frac{x - 6}{4})^{2} = 1 :

Up-down Hyperbola with

(h, k) = (6, 4), a = 4, b = 4

(6, 4 + 4), (6, 4 - 4)

Refine

(6, 8), (6, 0)

4 x^{2} - 9 y^{2} - 16 x - 72 y - 129 = 0

\frac{x ^{2}}{5} - \frac{y ^{2}}{7} = 1

- \frac{x ^{2}}{144} + \frac{y ^{2}}{25} = 1

16 x^{2} - 9 y^{2} - 64 x + 54 y - 161 = 0

y^{2} - x^{2} = \frac{4}{25}

What is the vertices ((y-4)/4)^2-((x-6)/4)^2=1 ?
The vertices ((y-4)/4)^2-((x-6)/4)^2=1 is (6,8),(6,0)