What is the center x^2+36y^2-16x-72y+64=0 ?

The center x^2+36y^2-16x-72y+64=0 is (8,1)

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center x^2+36y^2-16x-72y+64=0

center

x^{2} + 36 y^{2} - 16 x - 72 y + 64 = 0

(8, 1)

Solution steps

x^{2} + 36 y^{2} - 16 x - 72 y + 64 = 0

Rewrite

x^{2} + 36 y^{2} - 16 x - 72 y + 64 = 0

in the form of the standard ellipse equation

\frac{( x - 8 ) ^{2}}{6 ^{2}} + \frac{( y - 1 ) ^{2}}{1 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (8, 1), a = 6, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (8, 1), a = 6, b = 1

And the center is:

(8, 1)

\frac{x ^{2}}{49} + \frac{y ^{2}}{16} = 1

9 X^{2} + 4 Y^{2} = 36

\frac{x ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1

9 x^{2} + 4 y^{2} + 18 x - 8 y - 131 = 0

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

What is the center x^2+36y^2-16x-72y+64=0 ?
The center x^2+36y^2-16x-72y+64=0 is (8,1)