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

The vertices (x^2)/9-(y^2)/(49)=1 is (3,0),(-3,0)

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

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, 0), a = 3, b = 7

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

Refine

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

4 x^{2} - \frac{1}{2} y^{2} + 8 x + 2 y = 2

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 1 ) ^{2}}{49} = 1

\frac{y ^{2}}{9} - \frac{x ^{2}}{4} = 1

\frac{x ^{2}}{100} - \frac{y ^{2}}{64} = 1

\frac{y ^{2}}{25} - \frac{x ^{2}}{81} = 1

What is the vertices (x^2)/9-(y^2)/(49)=1 ?
The vertices (x^2)/9-(y^2)/(49)=1 is (3,0),(-3,0)