What is the vertices 4x^2-8y^2=3 ?

The vertices 4x^2-8y^2=3 is ((sqrt(3))/2 ,0),(-(sqrt(3))/2 ,0)

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vertices 4x^2-8y^2=3

vertices

4 x^{2} - 8 y^{2} = 3

(\frac{3}{2}, 0), (- \frac{3}{2}, 0)

Solution steps

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

Calculate Hyperbola properties

4 x^{2} - 8 y^{2} = 3 :

Right-left Hyperbola with

(h, k) = (0, 0), a = \frac{3}{2}, b = \frac{3}{2 2}

(0 + \frac{3}{2}, 0), (0 - \frac{3}{2}, 0)

Refine

(\frac{3}{2}, 0), (- \frac{3}{2}, 0)

2 x^{2} - 3 y^{2} = 6

\frac{( x + 3 ) ^{2}}{4} - \frac{( y - 1 ) ^{2}}{16} = 1

(\frac{y - 4}{16})^{2} - (\frac{x - 6}{16})^{2} = 1

\frac{( x - 1 ) ^{2}}{25} - \frac{( y + 6 ) ^{2}}{49} = 1

4 x^{2} - 24 x - 36 y^{2} - 360 y + 864 = 0

What is the vertices 4x^2-8y^2=3 ?
The vertices 4x^2-8y^2=3 is ((sqrt(3))/2 ,0),(-(sqrt(3))/2 ,0)