What is the center 5x^2+9y^2-10x+4=0 ?

The center 5x^2+9y^2-10x+4=0 is (1,0)

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

center

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

(1, 0)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, 0), a = \frac{1}{5}, b = \frac{1}{3}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, 0), a = \frac{1}{5}, b = \frac{1}{3}

And the center is:

(1, 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

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