What is the vertices (x^2)/7-(y^2)/6 =1 ?

The vertices (x^2)/7-(y^2)/6 =1 is (sqrt(7),0),(-sqrt(7),0)

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vertices (x^2)/7-(y^2)/6 =1

vertices

\frac{x ^{2}}{7} - \frac{y ^{2}}{6} = 1

(7, 0), (- 7, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{7} - \frac{y ^{2}}{6} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 7, b = 6

(0 + 7, 0), (0 - 7, 0)

Refine

(7, 0), (- 7, 0)

9 x^{2} - 16 y^{2} + 54 x - 16 y = - 113

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

4 x^{2} - 9 y^{2} - 16 x - 72 y - 129 = 0

\frac{x ^{2}}{5} - \frac{y ^{2}}{7} = 1

- \frac{x ^{2}}{144} + \frac{y ^{2}}{25} = 1

What is the vertices (x^2)/7-(y^2)/6 =1 ?
The vertices (x^2)/7-(y^2)/6 =1 is (sqrt(7),0),(-sqrt(7),0)