What is the vertices 9y^2-x^2+54y+2x+62=0 ?

The vertices 9y^2-x^2+54y+2x+62=0 is (1,-3+sqrt(2)),(1,-3-sqrt(2))

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vertices 9y^2-x^2+54y+2x+62=0

vertices

9 y^{2} - x^{2} + 54 y + 2 x + 62 = 0

(1, - 3 + 2), (1, - 3 - 2)

Solution steps

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

Calculate Hyperbola properties

9 y^{2} - x^{2} + 54 y + 2 x + 62 = 0 :

Up-down Hyperbola with

(h, k) = (1, - 3), a = 2, b = 32

(1, - 3 + 2), (1, - 3 - 2)

x^{2} - y^{2} = 6

4 x^{2} - y^{2} + 24 x - 4 y = 4

\frac{( x - 3 ) ^{2}}{3} - \frac{( y - 2 ) ^{2}}{22} = 1

16 x^{2} - 9 y^{2} + 64 x - 90 y = 305

4 x^{2} - 9 y^{2} - 36 < 0

What is the vertices 9y^2-x^2+54y+2x+62=0 ?
The vertices 9y^2-x^2+54y+2x+62=0 is (1,-3+sqrt(2)),(1,-3-sqrt(2))