What is the vertices x^2-y^2+8x-2y-10=0 ?

The vertices x^2-y^2+8x-2y-10=0 is (1,-1),(-9,-1)

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vertices x^2-y^2+8x-2y-10=0

vertices

x^{2} - y^{2} + 8 x - 2 y - 10 = 0

(1, - 1), (- 9, - 1)

Solution steps

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

Calculate Hyperbola properties

x^{2} - y^{2} + 8 x - 2 y - 10 = 0 :

Right-left Hyperbola with

(h, k) = (- 4, - 1), a = 5, b = 5

(- 4 + 5, - 1), (- 4 - 5, - 1)

Refine

(1, - 1), (- 9, - 1)

25 y^{2} - 9 x^{2} = 225

9 x^{2} - y^{2} = 36

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

\frac{x ^{2}}{4} - \frac{y ^{2}}{36} = 1

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

What is the vertices x^2-y^2+8x-2y-10=0 ?
The vertices x^2-y^2+8x-2y-10=0 is (1,-1),(-9,-1)