What is the center 3x^2+4y^2=12 ?

The center 3x^2+4y^2=12 is (0,0)

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

center

(0, 0)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{2 ^{2}} + \frac{( y - 0 ) ^{2}}{( 3 ) ^{2}} = 1

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 2, b = 3

And the center is:

(0, 0)