What is the vertices x^2-4y^2+2x+8y-7=0 ?

The vertices x^2-4y^2+2x+8y-7=0 is (1,1),(-3,1)

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vertices x^2-4y^2+2x+8y-7=0

vertices

x^{2} - 4 y^{2} + 2 x + 8 y - 7 = 0

(1, 1), (- 3, 1)

Solution steps

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

Calculate Hyperbola properties

x^{2} - 4 y^{2} + 2 x + 8 y - 7 = 0 :

Right-left Hyperbola with

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

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

Refine

(1, 1), (- 3, 1)

4 x^{2} - 25 y^{2} - 40 x - 150 y - 225 = 0

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

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

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

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

What is the vertices x^2-4y^2+2x+8y-7=0 ?
The vertices x^2-4y^2+2x+8y-7=0 is (1,1),(-3,1)