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

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

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

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Refine

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

16 x^{2} - 9 y^{2} - 64 x + 54 y - 161 = 0

y^{2} - x^{2} = \frac{4}{25}

- 16 x^{2} + 25 y^{2} - 64 x - 150 y - 239 = 0

- 9 x^{2} + 16 y^{2} - 90 x + 128 y - 113 = 0

2 y^{2} - 3 x^{2} - 4 y + 12 x + 8 = 0

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