What is the vertices 9+x^2-3y^2=0 ?

The vertices 9+x^2-3y^2=0 is (0,sqrt(3)),(0,-sqrt(3))

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vertices 9+x^2-3y^2=0

vertices

9 + x^{2} - 3 y^{2} = 0

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

Solution steps

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

Calculate Hyperbola properties

9 + x^{2} - 3 y^{2} = 0 :

Up-down Hyperbola with

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

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

Refine

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

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

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

\frac{y ^{2}}{36} - \frac{x ^{2}}{81} = 1

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

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

What is the vertices 9+x^2-3y^2=0 ?
The vertices 9+x^2-3y^2=0 is (0,sqrt(3)),(0,-sqrt(3))