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

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

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

Solution steps

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

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)

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

x^{2} + y^{2} = 100

y^{2} = - 14 x

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

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

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)