What is the center 4x^2+9y^2+32x-18y+37=0 ?

The center 4x^2+9y^2+32x-18y+37=0 is (-4,1)

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center 4x^2+9y^2+32x-18y+37=0

center

4 x^{2} + 9 y^{2} + 32 x - 18 y + 37 = 0

(- 4, 1)

Solution steps

4 x^{2} + 9 y^{2} + 32 x - 18 y + 37 = 0

Rewrite

4 x^{2} + 9 y^{2} + 32 x - 18 y + 37 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 4, 1), a = 3, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 4, 1), a = 3, b = 2

And the center is:

(- 4, 1)

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

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

12 x^{2} + 20 y^{2} - 12 x + 40 y - 37 = 0

6.25 x^{2} + 16 y^{2} - 12.5 x + 64 y - 29.75 = 0

2 x^{2} + y^{2} + 12 x - 6 y + 26 = 0

What is the center 4x^2+9y^2+32x-18y+37=0 ?
The center 4x^2+9y^2+32x-18y+37=0 is (-4,1)