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

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

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

vertices

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

(4, 2), (- 4, 2)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 2), a = 4, b = 3

(0 + 4, 2), (0 - 4, 2)

Refine

(4, 2), (- 4, 2)

x^{2} + 10 x + 9 y^{2} + 16 = 0

\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1

4 x^{2} + 9 y^{2} - 8 x - 36 y + 4 = 0

\frac{( x - 6 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{25} = 1

\frac{( x - 1 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{9} = 1

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