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

The vertices (x^2}{36}-\frac{y^2)/4 =1 is (6,0),(-6,0)

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vertices (x^2}{36}-\frac{y^2)/4 =1

vertices

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

(6, 0), (- 6, 0)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

(0 + 6, 0), (0 - 6, 0)

Refine

(6, 0), (- 6, 0)

\frac{x ^{2}}{4} - \frac{y ^{2}}{5} = 1

18 x^{2} - 32 y^{2} - 108 y - 128 x - 254 = 0

9 y^{2} - 25 x^{2} - 100 x - 108 y - 1 = 0

\frac{( x + 5 ) ^{2}}{4} - \frac{y ^{2}}{4} = 1

- 9 x^{2} + 8 y^{2} + 36 x + 48 y = 36

What is the vertices (x^2}{36}-\frac{y^2)/4 =1 ?
The vertices (x^2}{36}-\frac{y^2)/4 =1 is (6,0),(-6,0)