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

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

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center 2x^2+y^2-2x+16y+1=0

center

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

(\frac{1}{2}, - 8)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - \frac{1}{2} ) ^{2}}{( \frac{127}{2} ) ^{2}} + \frac{( y - ( - 8 ) ) ^{2}}{( \frac{127}{2} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (\frac{1}{2}, - 8), a = \frac{127}{2}, b = \frac{127}{2}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (\frac{1}{2}, - 8), b = \frac{127}{2}, a = \frac{127}{2}

And the center is:

(\frac{1}{2}, - 8)

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{3} = 1

25 x^{2} + 16 y^{2} - 150 x + 128 y + 81 = 0

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{4} = 1

x^{2} + y^{2} - 6 + 2 y + 6 = 0

2 x^{2} + 3 y^{2} + 4 x = 1

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