What is the vertices (y^2)/9-(x^2)/(16)=1 ?

The vertices (y^2)/9-(x^2)/(16)=1 is (0,3),(0,-3)

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vertices (y^2)/9-(x^2)/(16)=1

vertices

\frac{y ^{2}}{9} - \frac{x ^{2}}{16} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{y ^{2}}{9} - \frac{x ^{2}}{16} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 3, b = 4

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

Refine

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

2 x^{2} + 7 y^{2} = 14

49 y^{2} - x^{2} = 49

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

\frac{y ^{2}}{9} - \frac{x ^{2}}{16} = 1

f (x) = - x^{2} - 4 x - 3

What is the vertices (y^2)/9-(x^2)/(16)=1 ?
The vertices (y^2)/9-(x^2)/(16)=1 is (0,3),(0,-3)