What is the center 9x^2+4y^2+18x-8y-131=0 ?

The center 9x^2+4y^2+18x-8y-131=0 is (-1,1)

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

center

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

(- 1, 1)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{4 ^{2}} + \frac{( y - 1 ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, 1), a = 4, 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 = 4

And the center is:

(- 1, 1)

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

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

\frac{( x - 2 ) ^{2}}{36} + \frac{( y - 6 ) ^{2}}{4} = 1

\frac{( x - 5 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{4} = 1

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

What is the center 9x^2+4y^2+18x-8y-131=0 ?
The center 9x^2+4y^2+18x-8y-131=0 is (-1,1)