What is the vertices (x^2)/(144)-(y^2)/(25)=1 ?

The vertices (x^2)/(144)-(y^2)/(25)=1 is (12,0),(-12,0)

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vertices (x^2)/(144)-(y^2)/(25)=1

vertices

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

(12, 0), (- 12, 0)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, 0), a = 12, b = 5

(0 + 12, 0), (0 - 12, 0)

Refine

(12, 0), (- 12, 0)

9 y^{2} - x^{2} + 2 x + 54 y + 62 = 0

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 5 ) ^{2}}{9} = 1

9 x^{2} - 4 y^{2} - 54 x - 8 y + 113 = 0

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

\frac{x ^{2}}{64} - \frac{y ^{2}}{36} = 1

What is the vertices (x^2)/(144)-(y^2)/(25)=1 ?
The vertices (x^2)/(144)-(y^2)/(25)=1 is (12,0),(-12,0)