What is the vertices (x^2)/4-(y^2)/(36)=1 ?

The vertices (x^2)/4-(y^2)/(36)=1 is (2,0),(-2,0)

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vertices (x^2)/4-(y^2)/(36)=1

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, 0), a = 2, b = 6

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

Refine

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

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

\frac{1}{144} x^{2} - \frac{1}{196} y^{2} = 1

\frac{y ^{2}}{1 2 ^{2}} - \frac{x ^{2}}{5 ^{2}} = 1

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

64 x^{2} + 128 x - 9 y^{2} - 72 y - 656 = 0

What is the vertices (x^2)/4-(y^2)/(36)=1 ?
The vertices (x^2)/4-(y^2)/(36)=1 is (2,0),(-2,0)