What is the vertices x^2-3y^2+2x+6y-1=0 ?

The vertices x^2-3y^2+2x+6y-1=0 is (-1,1+sqrt(1/3)),(-1,1-sqrt(1/3))

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vertices x^2-3y^2+2x+6y-1=0

vertices

x^{2} - 3 y^{2} + 2 x + 6 y - 1 = 0

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

Solution steps

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

Calculate Hyperbola properties

x^{2} - 3 y^{2} + 2 x + 6 y - 1 = 0 :

Up-down Hyperbola with

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

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

9 x^{2} - 16 y^{2} - 18 x - 64 y - 199 = 0

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

\frac{( y + 4 ) ^{2}}{9} - \frac{( x + 2 ) ^{2}}{4} = 1

x^{2} - 3 y^{2} - 2 x - 6 y + 7 = 0

\frac{x ^{2}}{29} - \frac{y ^{2}}{20} = 1

What is the vertices x^2-3y^2+2x+6y-1=0 ?
The vertices x^2-3y^2+2x+6y-1=0 is (-1,1+sqrt(1/3)),(-1,1-sqrt(1/3))