What is the vertices (x^2)/(64)-(y^2)/(49)=1 ?

The vertices (x^2)/(64)-(y^2)/(49)=1 is (8,0),(-8,0)

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vertices (x^2)/(64)-(y^2)/(49)=1

vertices

\frac{x ^{2}}{64} - \frac{y ^{2}}{49} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{64} - \frac{y ^{2}}{49} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 8, b = 7

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

Refine

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

\frac{( y - 5 ) ^{2}}{4} - \frac{( x + 3 ) ^{2}}{9} = 1

9 x^{2} - 4 y^{2} + 54 x + 16 y + 66 = 0

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

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

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

What is the vertices (x^2)/(64)-(y^2)/(49)=1 ?
The vertices (x^2)/(64)-(y^2)/(49)=1 is (8,0),(-8,0)