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

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

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vertices 8y^2-4x^2=4

vertices

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

(0, \frac{1}{2}), (0, - \frac{1}{2})

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (0, 0), a = \frac{1}{2}, b = 1

(0, 0 + \frac{1}{2}), (0, 0 - \frac{1}{2})

Refine

(0, \frac{1}{2}), (0, - \frac{1}{2})

9 x^{2} - y^{2} + 18 x + 8 y = 43

\frac{( x - 7 ) ^{2}}{36} - \frac{( y - 3 ) ^{2}}{64} = 1

16 x^{2} - 3 y^{2} - 32 x + 12 y - 44 = 0

- 4 y^{2} + 9 x^{2} - 27 x - 16 y = 2

\frac{( y + 3 ) ^{2}}{4} - \frac{( x + 1 ) ^{2}}{49} = 1

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