What is the vertices (y^2)/4-(x^2)/5 =1 ?

The vertices (y^2)/4-(x^2)/5 =1 is (0,2),(0,-2)

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vertices (y^2)/4-(x^2)/5 =1

vertices

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

(0, 2), (0, - 2)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (0, 0), a = 2, b = 5

(0, 0 + 2), (0, 0 - 2)

Refine

(0, 2), (0, - 2)

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

\frac{x ^{2}}{100} + \frac{y ^{2}}{64} = 1

4 x^{2} - 20 x - 24 y + 97 = 0

x^{2} + y^{2} = 16

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

What is the vertices (y^2)/4-(x^2)/5 =1 ?
The vertices (y^2)/4-(x^2)/5 =1 is (0,2),(0,-2)