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

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

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vertices x^2+(y^2)/(16)=1

vertices

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), b = 4, a = 1

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

Refine

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

y^{2} + 2 x^{2} = 1

\frac{( y + 3 ) ^{2}}{9} - \frac{( x - 2 ) ^{2}}{16} = 1

y^{2} - 10 y - 12 x + 37 = 0

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

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

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