What is the vertices 4x^2-9y^2-16x-72y-129=0 ?

The vertices 4x^2-9y^2-16x-72y-129=0 is (5/2 ,-4),(3/2 ,-4)

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

vertices

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

(\frac{5}{2}, - 4), (\frac{3}{2}, - 4)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (2, - 4), a = \frac{1}{2}, b = \frac{1}{3}

(2 + \frac{1}{2}, - 4), (2 - \frac{1}{2}, - 4)

Refine

(\frac{5}{2}, - 4), (\frac{3}{2}, - 4)

\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}

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

What is the vertices 4x^2-9y^2-16x-72y-129=0 ?
The vertices 4x^2-9y^2-16x-72y-129=0 is (5/2 ,-4),(3/2 ,-4)