What is the vertices 9x^2-4y^2+54x+16y+66=0 ?

The vertices 9x^2-4y^2+54x+16y+66=0 is (-3, 5/2),(-3, 3/2)

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vertices 9x^2-4y^2+54x+16y+66=0

vertices

9 x^{2} - 4 y^{2} + 54 x + 16 y + 66 = 0

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

Solution steps

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

Calculate Hyperbola properties

9 x^{2} - 4 y^{2} + 54 x + 16 y + 66 = 0 :

Up-down Hyperbola with

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

(- 3, 2 + \frac{1}{2}), (- 3, 2 - \frac{1}{2})

Refine

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

\frac{y ^{2}}{36} - \frac{x ^{2}}{9} = 1

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

\frac{( y + 4 ) ^{2}}{4} - \frac{( x + 4 ) ^{2}}{25} = 1

4 x^{2} - 5 y^{2} + 20 = 0

9 (x - 3)^{2} - 4 (y - 1)^{2} = - 36

What is the vertices 9x^2-4y^2+54x+16y+66=0 ?
The vertices 9x^2-4y^2+54x+16y+66=0 is (-3, 5/2),(-3, 3/2)