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

The vertices (y^2)/(25)-(x^2)/(144)=1 is (0,5),(0,-5)

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

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Refine

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

4 x^{2} - 9 y^{2} - 24 x - 36 y - 36 = 0

9 x^{2} - 25 y^{2} = 225

\frac{x ^{2}}{36} - \frac{y ^{2}}{4} = 1

\frac{x ^{2}}{4} - \frac{y ^{2}}{5} = 1

18 x^{2} - 32 y^{2} - 108 y - 128 x - 254 = 0

What is the vertices (y^2)/(25)-(x^2)/(144)=1 ?
The vertices (y^2)/(25)-(x^2)/(144)=1 is (0,5),(0,-5)