What is the center 4x^2+7y^2<28 ?

The center 4x^2+7y^2<28 is (0,0)

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center 4x^2+7y^2<28

center

4 x^{2} + 7 y^{2} < 28

(0, 0)

Solution steps

4 x^{2} + 7 y^{2} < 28

Rewrite

4 x^{2} + 7 y^{2} < 28

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( 7 ) ^{2}} + \frac{( y - 0 ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = 7, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 7, b = 2

And the center is:

(0, 0)

\frac{( x + 4 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{16} = 1

16 x^{2} + 25 y^{2} - 64 x + 150 y - 11 = 0

\frac{( x - 2 ) ^{2}}{64} + \frac{( y + 4 ) ^{2}}{16} = 1

x^{2} + \frac{y ^{2}}{9} = 1

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