What is the center 9x^2+16y^2-18x-64y-71=0 ?

The center 9x^2+16y^2-18x-64y-71=0 is (1,2)

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center 9x^2+16y^2-18x-64y-71=0

center

9 x^{2} + 16 y^{2} - 18 x - 64 y - 71 = 0

(1, 2)

Solution steps

9 x^{2} + 16 y^{2} - 18 x - 64 y - 71 = 0

Rewrite

9 x^{2} + 16 y^{2} - 18 x - 64 y - 71 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, 2), a = 4, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, 2), a = 4, b = 3

And the center is:

(1, 2)

3 x^{2} + 4 y^{2} + 24 x - 16 y + 52 = 0

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

9 x^{2} + 4 y^{2} - 18 x + 8 y - 23 = 0

\frac{x ^{2}}{9} + \frac{y ^{2}}{36} = 1

25 x^{2} + 9 y^{2} - 250 x - 90 y + 625 = 0

What is the center 9x^2+16y^2-18x-64y-71=0 ?
The center 9x^2+16y^2-18x-64y-71=0 is (1,2)