What is the vertices (x^2)/4-(y^2)/4 =1 ?

The vertices (x^2)/4-(y^2)/4 =1 is (2,0),(-2,0)

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vertices (x^2)/4-(y^2)/4 =1

vertices

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

(2, 0), (- 2, 0)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, 0), a = 2, b = 2

(0 + 2, 0), (0 - 2, 0)

Refine

(2, 0), (- 2, 0)

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

4 x^{2} - 25 y^{2} + 16 x + 50 y - 109 = 0

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

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

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

What is the vertices (x^2)/4-(y^2)/4 =1 ?
The vertices (x^2)/4-(y^2)/4 =1 is (2,0),(-2,0)