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Popular >

vertices 3x^2+2y^2=6

Solution

vertices

3 x^{2} + 2 y^{2} = 6

Solution

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

Solution steps

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

Calculate ellipse properties

3 x^{2} + 2 y^{2} = 6 :

Ellipse with center

(h, k) = (0, 0), b = 3, a = 2

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

Refine

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

Popular Examples

foci (x^2)/(25)+(y^2)/(25)=1 foci

\frac{x ^{2}}{25} + \frac{y ^{2}}{25} = 1

y^2+20x+8y+56=0

y^{2} + 20 x + 8 y + 56 = 0

x^{2} + y^{2} \leq 16

vertices f(x)=x^2+4x vertices

f (x) = x^{2} + 4 x

foci (x^2)/4+(y^2)/(25)=1 foci

\frac{x ^{2}}{4} + \frac{y ^{2}}{25} = 1

Frequently Asked Questions (FAQ)

What is the vertices 3x^2+2y^2=6 ?
The vertices 3x^2+2y^2=6 is (0,sqrt(3)),(0,-sqrt(3))