What is the vertices (x^2}{64}-\frac{y^2)/9 =1 ?

The vertices (x^2}{64}-\frac{y^2)/9 =1 is (8,0),(-8,0)

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vertices (x^2}{64}-\frac{y^2)/9 =1

vertices

\frac{x ^{2}}{64} - \frac{y ^{2}}{9} = 1

(8, 0), (- 8, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{64} - \frac{y ^{2}}{9} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 8, b = 3

(0 + 8, 0), (0 - 8, 0)

Refine

(8, 0), (- 8, 0)

\frac{y ^{2}}{16} - \frac{x ^{2}}{25} = 1

\frac{x ^{2}}{64} - \frac{y ^{2}}{36} = 1

9 x^{2} - 4 y^{2} - 90 x - 32 y = - 305

36 x^{2} + 288 x + 64 y^{2} - 256 y = 1472

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

What is the vertices (x^2}{64}-\frac{y^2)/9 =1 ?
The vertices (x^2}{64}-\frac{y^2)/9 =1 is (8,0),(-8,0)