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

The vertices ((x+2)^2}{16}+\frac{y^2)/4 =1 is (2,0),(-6,0)

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vertices ((x+2)^2}{16}+\frac{y^2)/4 =1

vertices

\frac{( x + 2 ) ^{2}}{16} + \frac{y ^{2}}{4} = 1

(2, 0), (- 6, 0)

Solution steps

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

Calculate ellipse properties

\frac{( x + 2 ) ^{2}}{16} + \frac{y ^{2}}{4} = 1 :

Ellipse with center

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

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

Refine

(2, 0), (- 6, 0)

\frac{x ^{2}}{36} + \frac{y ^{2}}{11} = 1

\frac{( x + 4 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{16} = 1

5 x^{2} + 9 y^{2} - 20 x + 54 y + 56 = 0

\frac{x ^{2}}{5} + \frac{y ^{2}}{7} = 1

9 x^{2} + 36 y^{2} = 324

What is the vertices ((x+2)^2}{16}+\frac{y^2)/4 =1 ?
The vertices ((x+2)^2}{16}+\frac{y^2)/4 =1 is (2,0),(-6,0)