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

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

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

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Refine

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

x^{2} + 4 y^{2} - 6 x + 16 y + 21 = 0

x^{2} + y^{2} - 6 x + 4 y + 9 = 0

f (x) = x^{2} - 4 x + 3

\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1

x^{2} + y^{2} + 4 x - 6 y - 3 = 0

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