What is the vertices 4x^2-9y^2-16x+18y-7=0 ?

The vertices 4x^2-9y^2-16x+18y-7=0 is (2+sqrt(7/2),1),(2-sqrt(7/2),1)

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vertices 4x^2-9y^2-16x+18y-7=0

vertices

4 x^{2} - 9 y^{2} - 16 x + 18 y - 7 = 0

(2 + \frac{7}{2}, 1), (2 - \frac{7}{2}, 1)

Solution steps

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

Calculate Hyperbola properties

4 x^{2} - 9 y^{2} - 16 x + 18 y - 7 = 0 :

Right-left Hyperbola with

(h, k) = (2, 1), a = \frac{7}{2}, b = \frac{14}{3}

(2 + \frac{7}{2}, 1), (2 - \frac{7}{2}, 1)

\frac{x ^{2}}{7} - \frac{y ^{2}}{6} = 1

9 x^{2} - 16 y^{2} + 54 x - 16 y = - 113

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

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

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

What is the vertices 4x^2-9y^2-16x+18y-7=0 ?
The vertices 4x^2-9y^2-16x+18y-7=0 is (2+sqrt(7/2),1),(2-sqrt(7/2),1)