What is the vertices 9x^2-4y^2-54x-8y+113=0 ?

The vertices 9x^2-4y^2-54x-8y+113=0 is (3,2),(3,-4)

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vertices 9x^2-4y^2-54x-8y+113=0

vertices

9 x^{2} - 4 y^{2} - 54 x - 8 y + 113 = 0

(3, 2), (3, - 4)

Solution steps

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

Calculate Hyperbola properties

9 x^{2} - 4 y^{2} - 54 x - 8 y + 113 = 0 :

Up-down Hyperbola with

(h, k) = (3, - 1), a = 3, b = 2

(3, - 1 + 3), (3, - 1 - 3)

Refine

(3, 2), (3, - 4)

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

\frac{x ^{2}}{64} - \frac{y ^{2}}{36} = 1

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

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

4 x^{2} - 9 y^{2} - 24 x - 36 y - 36 = 0

What is the vertices 9x^2-4y^2-54x-8y+113=0 ?
The vertices 9x^2-4y^2-54x-8y+113=0 is (3,2),(3,-4)