What is the vertices (x^2)/5-(y^2)/7 =1 ?

The vertices (x^2)/5-(y^2)/7 =1 is (sqrt(5),0),(-sqrt(5),0)

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vertices (x^2)/5-(y^2)/7 =1

vertices

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

(5, 0), (- 5, 0)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, 0), a = 5, b = 7

(0 + 5, 0), (0 - 5, 0)

Refine

(5, 0), (- 5, 0)

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

16 x^{2} - 9 y^{2} - 64 x + 54 y - 161 = 0

y^{2} - x^{2} = \frac{4}{25}

- 16 x^{2} + 25 y^{2} - 64 x - 150 y - 239 = 0

- 9 x^{2} + 16 y^{2} - 90 x + 128 y - 113 = 0

What is the vertices (x^2)/5-(y^2)/7 =1 ?
The vertices (x^2)/5-(y^2)/7 =1 is (sqrt(5),0),(-sqrt(5),0)