What is the vertices 4x^2-1/2 y^2+8x-2y=2 ?

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

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vertices 4x^2-1/2 y^2+8x-2y=2

vertices

4 x^{2} - \frac{1}{2} y^{2} + 8 x - 2 y = 2

(0, - 2), (- 2, - 2)

Solution steps

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

Calculate Hyperbola properties

4 x^{2} - \frac{1}{2} y^{2} + 8 x - 2 y = 2 :

Right-left Hyperbola with

(h, k) = (- 1, - 2), a = 1, b = 22

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

Refine

(0, - 2), (- 2, - 2)

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

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

\frac{y ^{2}}{225} - \frac{x ^{2}}{64} = 1

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

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

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