What is the center 9x^2+4y^2=25 ?

The center 9x^2+4y^2=25 is (0,0)

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center 9x^2+4y^2=25

center

9 x^{2} + 4 y^{2} = 25

(0, 0)

Solution steps

9 x^{2} + 4 y^{2} = 25

Rewrite

9 x^{2} + 4 y^{2} = 25

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{5}{3} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( \frac{5}{2} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = \frac{5}{3}, b = \frac{5}{2}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = \frac{5}{2}, a = \frac{5}{3}

And the center is:

(0, 0)

9 x^{2} + 25 y^{2} + 108 x - 200 y + 499 = 0

\frac{( x - 5 ) ^{2}}{49} + \frac{( y - 9 ) ^{2}}{25} = 1

4 x^{2} + 9 y^{2} = 25

3 x^{(2)} + 2 y^{(2)} - 12 x + 8 y + 19 = 0

\frac{x ^{2}}{9} + \frac{y ^{2}}{16} = 1