What is the vertices (x^2)/9+(y^2)/(16)>1 ?

The vertices (x^2)/9+(y^2)/(16)>1 is (0,4),(0,-4)

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vertices (x^2)/9+(y^2)/(16)>1

vertices

(0, 4), (0, - 4)

Solution steps

(h, k + b), (h, k - b)

Rewrite as

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

Calculate ellipse properties

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

Ellipse with center

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

(0, 0 + 4), (0, 0 - 4)

Refine

(0, 4), (0, - 4)

What is the vertices (x^2)/9+(y^2)/(16)>1 ?
The vertices (x^2)/9+(y^2)/(16)>1 is (0,4),(0,-4)