What is the vertices (x^2}{16}+\frac{y^2)/9 =1 ?

The vertices (x^2}{16}+\frac{y^2)/9 =1 is (4,0),(-4,0)

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vertices (x^2}{16}+\frac{y^2)/9 =1

vertices

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

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

x^{2} + y^{2} - 2 x + 6 y + 6 = 0

y = - x^{2} + 4 x + 1

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

\frac{y ^{2}}{64} - \frac{x ^{2}}{9} = 1

x^{2} + 4 y^{2} - 6 x + 16 y + 21 = 0

What is the vertices (x^2}{16}+\frac{y^2)/9 =1 ?
The vertices (x^2}{16}+\frac{y^2)/9 =1 is (4,0),(-4,0)