What is the vertices (x^2)/4-(y^2)/(12)=1 ?

The vertices (x^2)/4-(y^2)/(12)=1 is (2,0),(-2,0)

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vertices (x^2)/4-(y^2)/(12)=1

vertices

\frac{x ^{2}}{4} - \frac{y ^{2}}{12} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{4} - \frac{y ^{2}}{12} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 2, b = 23

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

Refine

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

9 y^{2} - 4 x^{2} - 36 y - 8 x = 4

4 x^{2} - 25 y^{2} = 100

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

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

+ \frac{x ^{2}}{4} - \frac{y ^{2}}{9} = 1

What is the vertices (x^2)/4-(y^2)/(12)=1 ?
The vertices (x^2)/4-(y^2)/(12)=1 is (2,0),(-2,0)