What is the center 9x^2+16y^2-54x+64y+1=0 ?

The center 9x^2+16y^2-54x+64y+1=0 is (3,-2)

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

center

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

(3, - 2)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{4 ^{2}} + \frac{( y - ( - 2 ) ) ^{2}}{3 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, - 2), a = 4, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, - 2), a = 4, b = 3

And the center is:

(3, - 2)

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

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

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

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

4 x^{2} + 9 y^{2} = 36

What is the center 9x^2+16y^2-54x+64y+1=0 ?
The center 9x^2+16y^2-54x+64y+1=0 is (3,-2)