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Popular >

vertices 3x^2+4y^2=12

Solution

vertices

3 x^{2} + 4 y^{2} = 12

Solution

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

Solution steps

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

Calculate ellipse properties

3 x^{2} + 4 y^{2} = 12 :

Ellipse with center

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

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

Refine

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

Popular Examples

vertices (x^2)/(6^2)+(y^2)/(7^2)=1 vertices

\frac{x ^{2}}{6 ^{2}} + \frac{y ^{2}}{7 ^{2}} = 1

vertices (x^2}{49}+\frac{y^2)/4 =1 vertices

\frac{x ^{2}}{49} + \frac{y ^{2}}{4} = 1

vertices 25x^2+9y^2-150x-18y+9=0 vertices

25 x^{2} + 9 y^{2} - 150 x - 18 y + 9 = 0

vertices (x^2+8(y+5)^2)=20 vertices

(x^{2} + 8 (y + 5)^{2}) = 20

vertices x^2+4y^2=4x+8 vertices

x^{2} + 4 y^{2} = 4 x + 8

Frequently Asked Questions (FAQ)

What is the vertices 3x^2+4y^2=12 ?
The vertices 3x^2+4y^2=12 is (2,0),(-2,0)