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

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

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

center

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

(2, 3)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

And the center is:

(2, 3)

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

25 x^{2} + y^{2} - 100 x + 8 y + 91 = 0

49 x^{2} + 625 y^{2} = 30625

4 x^{2} + 9 y^{2} - 24 x + 36 y + 36 = 0

x^{2} + 3 y^{2} + 2 x - 12 y + 10 = 0

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