What is the center (x^2)/4+y^2=1 ?

The center (x^2)/4+y^2=1 is (0,0)

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center (x^2)/4+y^2=1

center

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

(0, 0)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = 2, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 2, b = 1

And the center is:

(0, 0)

5 x^{2} + 9 y^{2} - 10 x + 4 = 0

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

\frac{x ^{2}}{7} + \frac{y ^{2}}{16} = 1

25 (x - 2)^{2} + 9 (y + 3)^{2} = 225

9 x^{2} + 25 y^{2} - 36 x - 50 y + 60 = 0