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

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

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center 4x^2+y^2=16

center

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

(0, 0)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

And the center is:

(0, 0)

\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1

16 (x - 1)^{2} + 9 (y + 2)^{2} = 144

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 4 ) ^{2}}{9} = 1

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

\frac{( x - 8 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{9} = 1