What is the vertices 9x^2-4y^2+36x+8y+33=0 ?

The vertices 9x^2-4y^2+36x+8y+33=0 is (-2, 3/2),(-2, 1/2)

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vertices 9x^2-4y^2+36x+8y+33=0

vertices

9 x^{2} - 4 y^{2} + 36 x + 8 y + 33 = 0

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

Solution steps

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

Calculate Hyperbola properties

9 x^{2} - 4 y^{2} + 36 x + 8 y + 33 = 0 :

Up-down Hyperbola with

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

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

Refine

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

- 4 x^{2} + 24 x + 16 y^{2} - 128 y + 156 = 0

y^{2} - 9 x^{2} = 4

\frac{( x - 3 ) ^{2}}{49} - \frac{( y - 2 ) ^{2}}{16} = 1

(\frac{x}{9})^{2} - (\frac{y}{49})^{2} = 1

4 x^{2} - 8 y^{2} = 3

What is the vertices 9x^2-4y^2+36x+8y+33=0 ?
The vertices 9x^2-4y^2+36x+8y+33=0 is (-2, 3/2),(-2, 1/2)