What is the center 3x^2+4y^2+12x-8y+9=0 ?

The center 3x^2+4y^2+12x-8y+9=0 is (-2,1)

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center 3x^2+4y^2+12x-8y+9=0

center

3 x^{2} + 4 y^{2} + 12 x - 8 y + 9 = 0

(- 2, 1)

Solution steps

3 x^{2} + 4 y^{2} + 12 x - 8 y + 9 = 0

Rewrite

3 x^{2} + 4 y^{2} + 12 x - 8 y + 9 = 0

in the form of the standard ellipse equation

\frac{( x - ( - 2 ) ) ^{2}}{( \frac{7}{3} ) ^{2}} + \frac{( y - 1 ) ^{2}}{( \frac{7}{2} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 2, 1), a = \frac{7}{3}, b = \frac{7}{2}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 2, 1), a = \frac{7}{3}, b = \frac{7}{2}

And the center is:

(- 2, 1)

\frac{( x + 1 ) ^{2}}{9} + \frac{( y + 3 ) ^{2}}{36} = 1

x^{2} + y^{2} \leq y

25 (x - 5)^{2} + 16 (y + 4)^{2} = 400

4 x^{2} + 25 y^{2} - 32 x + 200 y + 364 = 0

(x 3)^{2} + (y + 4)^{2} = 25

What is the center 3x^2+4y^2+12x-8y+9=0 ?
The center 3x^2+4y^2+12x-8y+9=0 is (-2,1)