What is the vertices (x^2)/(25)-(y^2)/(121)=1 ?

The vertices (x^2)/(25)-(y^2)/(121)=1 is (5,0),(-5,0)

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vertices (x^2)/(25)-(y^2)/(121)=1

vertices

\frac{x ^{2}}{25} - \frac{y ^{2}}{121} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{25} - \frac{y ^{2}}{121} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 5, b = 11

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

Refine

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

6 x^{2} - 8 y^{2} + 12 x - 32 y = 178

9 + x^{2} - 3 y^{2} = 0

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

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

\frac{y ^{2}}{36} - \frac{x ^{2}}{81} = 1

What is the vertices (x^2)/(25)-(y^2)/(121)=1 ?
The vertices (x^2)/(25)-(y^2)/(121)=1 is (5,0),(-5,0)