What is the center 4x^2+2y^2+2x-8y+7=0 ?

The center 4x^2+2y^2+2x-8y+7=0 is (-1/4 ,2)

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center 4x^2+2y^2+2x-8y+7=0

center

4 x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

(- \frac{1}{4}, 2)

Solution steps

4 x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

Rewrite

4 x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- \frac{1}{4}, 2), a = \frac{5}{4}, b = \frac{5}{2 2}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- \frac{1}{4}, 2), b = \frac{5}{2 2}, a = \frac{5}{4}

And the center is:

(- \frac{1}{4}, 2)

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{81} = 1

\frac{( x - 1 ) ^{2}}{36} + \frac{( y + 2 ) ^{2}}{25} = 1

25 x^{2} + 4 y^{2} - 50 x + 16 y - 59 = 0

\frac{( x - 2 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{9} = 1

\frac{( x + 2 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} = 1

What is the center 4x^2+2y^2+2x-8y+7=0 ?
The center 4x^2+2y^2+2x-8y+7=0 is (-1/4 ,2)