What is the vertices x^2-3y^2-2x-6y+7=0 ?

The vertices x^2-3y^2-2x-6y+7=0 is (1,-1+sqrt(3)),(1,-1-sqrt(3))

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vertices x^2-3y^2-2x-6y+7=0

vertices

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

(1, - 1 + 3), (1, - 1 - 3)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (1, - 1), a = 3, b = 3

(1, - 1 + 3), (1, - 1 - 3)

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

y^{2} - x^{2} = \frac{9}{25}

16 y^{2} - 9 x^{2} + 36 x + 96 y - 36 = 0

- 25 x^{2} + 4 y^{2} = 100

\frac{y ^{2}}{144} - \frac{x ^{2}}{25} = 1

What is the vertices x^2-3y^2-2x-6y+7=0 ?
The vertices x^2-3y^2-2x-6y+7=0 is (1,-1+sqrt(3)),(1,-1-sqrt(3))