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Popular >

vertices 2x^2+y^2=4

Solution

vertices

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

Solution

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

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

2 x^{2} + y^{2} = 4 :

Ellipse with center

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

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

Refine

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

Popular Examples

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9 x^{2} + 4 y^{2} - 18 x - 16 y - 11 = 0

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

vertices ((x+3)^2)/(26)+((y-4)^2)/(16)=1 vertices

\frac{( x + 3 ) ^{2}}{26} + \frac{( y - 4 ) ^{2}}{16} = 1

vertices (x^2)/8+(y^2)/4 =1 vertices

\frac{x ^{2}}{8} + \frac{y ^{2}}{4} = 1

vertices 4x^2+y^2-32x+16y+124=0 vertices

4 x^{2} + y^{2} - 32 x + 16 y + 124 = 0

Frequently Asked Questions (FAQ)

What is the vertices 2x^2+y^2=4 ?
The vertices 2x^2+y^2=4 is (0,2),(0,-2)