What is the vertices 2x^2+7y^2=14 ?

The vertices 2x^2+7y^2=14 is (sqrt(7),0),(-sqrt(7),0)

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vertices 2x^2+7y^2=14

vertices

2 x^{2} + 7 y^{2} = 14

(7, 0), (- 7, 0)

Solution steps

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

Calculate ellipse properties

2 x^{2} + 7 y^{2} = 14 :

Ellipse with center

(h, k) = (0, 0), a = 7, b = 2

(0 + 7, 0), (0 - 7, 0)

Refine

(7, 0), (- 7, 0)

16 x^{2} + 25 y^{2} - 1600 = 0

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

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

What is the vertices 2x^2+7y^2=14 ?
The vertices 2x^2+7y^2=14 is (sqrt(7),0),(-sqrt(7),0)