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

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

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

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, 0), a = 3, b = 1

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

Refine

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

y^{2} - x^{2} = 1

\frac{( x - 5 ) ^{2}}{81} - \frac{( y - 7 ) ^{2}}{49} = 1

\frac{x ^{2}}{25} - \frac{y ^{2}}{64} = 1

x^{2} - 4 y^{2} = 16

9 x^{2} - y^{2} - 54 x - 2 y + 71 = 0

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