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

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

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

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Refine

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

\frac{( x - 3 ) ^{2}}{9} - \frac{( y - 2 ) ^{2}}{4} = 1

\frac{( x + 1 ) ^{2}}{9} - \frac{( y - 1 ) ^{2}}{8} = 1

\frac{x ^{2}}{60} - \frac{y ^{2}}{40} = 1

\frac{x ^{2}}{36} - \frac{y ^{2}}{25} = 1

x^{2} - 9 y^{2} = 36

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