What is the vertices (x^2}{25}-\frac{y^2)/4 =1 ?

The vertices (x^2}{25}-\frac{y^2)/4 =1 is (5,0),(-5,0)

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

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{25} - \frac{y ^{2}}{4} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 5, b = 2

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

Refine

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

x^{2} - 4 y^{2} + 2 x + 8 y - 7 = 0

4 x^{2} - 25 y^{2} - 40 x - 150 y - 225 = 0

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

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

4 (x - 2)^{2} - 9 (y + 4)^{2} = 1

What is the vertices (x^2}{25}-\frac{y^2)/4 =1 ?
The vertices (x^2}{25}-\frac{y^2)/4 =1 is (5,0),(-5,0)