What is the vertices (y^2)/9-(x^2)/4 =1 ?

The vertices (y^2)/9-(x^2)/4 =1 is (0,3),(0,-3)

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vertices (y^2)/9-(x^2)/4 =1

vertices

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

(0, 3), (0, - 3)

Solution steps

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

Calculate Hyperbola properties

\frac{y ^{2}}{9} - \frac{x ^{2}}{4} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 3, b = 2

(0, 0 + 3), (0, 0 - 3)

Refine

(0, 3), (0, - 3)

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

\frac{y ^{2}}{25} - \frac{x ^{2}}{81} = 1

9 x^{2} - 18 x - 16 y^{2} - 64 y = 199

\frac{y ^{2}}{49} - \frac{x ^{2}}{16} = 1

- 3 x^{2} + 2 y^{2} + 6 x - 7 y = 7

What is the vertices (y^2)/9-(x^2)/4 =1 ?
The vertices (y^2)/9-(x^2)/4 =1 is (0,3),(0,-3)