What is the center x^2+4y^2=16 ?

The center x^2+4y^2=16 is (0,0)

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center x^2+4y^2=16

center

x^{2} + 4 y^{2} = 16

(0, 0)

Solution steps

x^{2} + 4 y^{2} = 16

Rewrite

x^{2} + 4 y^{2} = 16

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{4 ^{2}} + \frac{( y - 0 ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = 4, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 4, b = 2

And the center is:

(0, 0)

\frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1

4 x^{2} + y^{2} = 1

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

3 x^{2} + 7 y^{2} - 12 x - 28 y + 19 = 0

x^{2} + 4 y^{2} - 6 x + 16 y + 21 = 0