What is the center 4x^2+y^2+8x+2y-31=0 ?

The center 4x^2+y^2+8x+2y-31=0 is (-1,-1)

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center 4x^2+y^2+8x+2y-31=0

center

4 x^{2} + y^{2} + 8 x + 2 y - 31 = 0

(- 1, - 1)

Solution steps

4 x^{2} + y^{2} + 8 x + 2 y - 31 = 0

Rewrite

4 x^{2} + y^{2} + 8 x + 2 y - 31 = 0

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{3 ^{2}} + \frac{( y - ( - 1 ) ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, - 1), a = 3, b = 6

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, - 1), b = 6, a = 3

And the center is:

(- 1, - 1)

\frac{( x + 5 ) ^{2}}{49} + \frac{( y - 2 ) ^{2}}{81} = 1

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{36} = 1

\frac{x ^{2}}{4} + y^{2} = 1

5 x^{2} + 9 y^{2} - 10 x + 4 = 0

3 x^{2} + 4 y^{2} = 12

What is the center 4x^2+y^2+8x+2y-31=0 ?
The center 4x^2+y^2+8x+2y-31=0 is (-1,-1)