What is the vertices (x^2)/4+y^2=1 ?

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

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vertices (x^2)/4+y^2=1

vertices

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), a = 2, b = 1

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

Refine

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

9 x^{2} + 4 y^{2} + 18 x - 8 y - 131 = 0

25 x^{2} - 200 x + y^{2} + 4 y + 379 = 0

4 x^{2} + 9 y^{2} + 32 x - 18 y + 37 = 0

\frac{( x + 1 ) ^{2}}{3} + \frac{( y - 1 ) ^{2}}{6} = 4

(x + 3)^{2} + y^{2} = 36