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)

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 1 ) ^{2}}{49} = 1

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

\frac{x ^{2}}{100} - \frac{y ^{2}}{64} = 1

\frac{y ^{2}}{25} - \frac{x ^{2}}{81} = 1

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

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)