What is the center x^2+4y^2-6x+8y+9=0 ?

The center x^2+4y^2-6x+8y+9=0 is (3,-1)

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center x^2+4y^2-6x+8y+9=0

center

x^{2} + 4 y^{2} - 6 x + 8 y + 9 = 0

(3, - 1)

Solution steps

x^{2} + 4 y^{2} - 6 x + 8 y + 9 = 0

Rewrite

x^{2} + 4 y^{2} - 6 x + 8 y + 9 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 1), a = 2, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, - 1), a = 2, b = 1

And the center is:

(3, - 1)

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

\frac{x ^{2}}{64} + \frac{y ^{2}}{100} = 1

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 5 ) ^{2}}{9} = 1

9 x^{2} + 16 y^{2} - 54 x + 64 y + 1 = 0

9 x^{2} + 16 y^{2} - 72 x + 64 y - 368 = 0