What is the vertices (x/9)^2-(y/(49))^2=1 ?

The vertices (x/9)^2-(y/(49))^2=1 is (9,0),(-9,0)

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vertices (x/9)^2-(y/(49))^2=1

vertices

(\frac{x}{9})^{2} - (\frac{y}{49})^{2} = 1

(9, 0), (- 9, 0)

Solution steps

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

Calculate Hyperbola properties

(\frac{x}{9})^{2} - (\frac{y}{49})^{2} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 9, b = 49

(0 + 9, 0), (0 - 9, 0)

Refine

(9, 0), (- 9, 0)

4 x^{2} - 8 y^{2} = 3

2 x^{2} - 3 y^{2} = 6

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

(\frac{y - 4}{16})^{2} - (\frac{x - 6}{16})^{2} = 1

\frac{( x - 1 ) ^{2}}{25} - \frac{( y + 6 ) ^{2}}{49} = 1

What is the vertices (x/9)^2-(y/(49))^2=1 ?
The vertices (x/9)^2-(y/(49))^2=1 is (9,0),(-9,0)