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

(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)

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)