What is the vertices (x^2)/9-(y^2)/(25)=1 ?

The vertices (x^2)/9-(y^2)/(25)=1 is (3,0),(-3,0)

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vertices (x^2)/9-(y^2)/(25)=1

vertices

\frac{x ^{2}}{9} - \frac{y ^{2}}{25} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{9} - \frac{y ^{2}}{25} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 3, b = 5

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

Refine

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

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

y = 2 x^{2}

\frac{y ^{2}}{9} - \frac{x ^{2}}{16} = 1

2 x^{2} + 7 y^{2} = 14

49 y^{2} - x^{2} = 49

What is the vertices (x^2)/9-(y^2)/(25)=1 ?
The vertices (x^2)/9-(y^2)/(25)=1 is (3,0),(-3,0)