What is the vertices (x^2)/9-(y^2)/(36)=1 ?

The vertices (x^2)/9-(y^2)/(36)=1 is (3,0),(-3,0)

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vertices (x^2)/9-(y^2)/(36)=1

vertices

\frac{x ^{2}}{9} - \frac{y ^{2}}{36} = 1

(3, 0), (- 3, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{9} - \frac{y ^{2}}{36} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 3, b = 6

(0 + 3, 0), (0 - 3, 0)

Refine

(3, 0), (- 3, 0)

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

2 x^{2} - 2 y^{2} - 20 x + 16 y - 16 = 0

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

\frac{( x - 1 ) ^{2}}{4} - \frac{( y + 3 ) ^{2}}{16} = 1

\frac{( y + 1 ) ^{2}}{25} - \frac{( x + 4 ) ^{2}}{64} = 1

What is the vertices (x^2)/9-(y^2)/(36)=1 ?
The vertices (x^2)/9-(y^2)/(36)=1 is (3,0),(-3,0)