What is the vertices x^2-4y^2+8x+16y-4=0 ?

The vertices x^2-4y^2+8x+16y-4=0 is (-2,2),(-6,2)

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vertices x^2-4y^2+8x+16y-4=0

vertices

x^{2} - 4 y^{2} + 8 x + 16 y - 4 = 0

(- 2, 2), (- 6, 2)

Solution steps

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

Calculate Hyperbola properties

x^{2} - 4 y^{2} + 8 x + 16 y - 4 = 0 :

Right-left Hyperbola with

(h, k) = (- 4, 2), a = 2, b = 1

(- 4 + 2, 2), (- 4 - 2, 2)

Refine

(- 2, 2), (- 6, 2)

9 y^{2} - 4 x^{2} + 32 x - 36 y - 64 = 0

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

9 (x - 1)^{2} - 16 (y + 2)^{2} = 144

9 x^{2} - 4 y^{2} + 54 x + 16 y + 29 = 0

\frac{( y + 3 ) ^{2}}{9} - \frac{( x - 2 ) ^{2}}{16} = 1

What is the vertices x^2-4y^2+8x+16y-4=0 ?
The vertices x^2-4y^2+8x+16y-4=0 is (-2,2),(-6,2)