What is the vertices 16x^2-9y^2-64x+54y-161=0 ?

The vertices 16x^2-9y^2-64x+54y-161=0 is (5,3),(-1,3)

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

vertices

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

(5, 3), (- 1, 3)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (2, 3), a = 3, b = 4

(2 + 3, 3), (2 - 3, 3)

Refine

(5, 3), (- 1, 3)

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

- 16 x^{2} + 25 y^{2} - 64 x - 150 y - 239 = 0

- 9 x^{2} + 16 y^{2} - 90 x + 128 y - 113 = 0

2 y^{2} - 3 x^{2} - 4 y + 12 x + 8 = 0

\frac{( y + 3 ) ^{2}}{9} - \frac{( x - 1 ) ^{2}}{18} = 1

What is the vertices 16x^2-9y^2-64x+54y-161=0 ?
The vertices 16x^2-9y^2-64x+54y-161=0 is (5,3),(-1,3)