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Popular >

vertices x^2+2y^2=3

Solution

vertices

x^{2} + 2 y^{2} = 3

Solution

(3, 0), (- 3, 0)

Solution steps

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

Calculate ellipse properties

x^{2} + 2 y^{2} = 3 :

Ellipse with center

(h, k) = (0, 0), a = 3, b = \frac{3}{2}

(0 + 3, 0), (0 - 3, 0)

Refine

(3, 0), (- 3, 0)

Popular Examples

vertices 25x^2+4y^2-50x+40y+25=0 vertices

25 x^{2} + 4 y^{2} - 50 x + 40 y + 25 = 0

vertices 9x^2+y^2+18x+4y+4=0 vertices

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

vertices (x^2)/9+(y^2)/(36)=1 vertices

\frac{x ^{2}}{9} + \frac{y ^{2}}{36} = 1

vertices 16x^2+25y^2=1600 vertices

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

vertices ((x+1)^2)/9+((y-3)^2)/(16)=1 vertices

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

Frequently Asked Questions (FAQ)

What is the vertices x^2+2y^2=3 ?
The vertices x^2+2y^2=3 is (sqrt(3),0),(-sqrt(3),0)