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

The vertices x^2-y^2+2x+2y-16=0 is (3,1),(-5,1)

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

vertices

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

(3, 1), (- 5, 1)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Refine

(3, 1), (- 5, 1)

8 x^{2} - 2 y^{2} = 16

\frac{( y - 2 ) ^{2}}{25} - \frac{x ^{2}}{9} = 1

- 4 x^{2} + y^{2} + 8 y - 20 = 0

\frac{y ^{2}}{25} - \frac{x ^{2}}{39} = 1

y^{2} - 9 x^{2} - 8 y - 54 x - 81 = 0

What is the vertices x^2-y^2+2x+2y-16=0 ?
The vertices x^2-y^2+2x+2y-16=0 is (3,1),(-5,1)