What is the center 3x^2+4y^2+24x-16y+52=0 ?

The center 3x^2+4y^2+24x-16y+52=0 is (-4,2)

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center 3x^2+4y^2+24x-16y+52=0

center

3 x^{2} + 4 y^{2} + 24 x - 16 y + 52 = 0

(- 4, 2)

Solution steps

3 x^{2} + 4 y^{2} + 24 x - 16 y + 52 = 0

Rewrite

3 x^{2} + 4 y^{2} + 24 x - 16 y + 52 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

And the center is:

(- 4, 2)

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

9 x^{2} + 4 y^{2} - 18 x + 8 y - 23 = 0

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

25 x^{2} + 9 y^{2} - 250 x - 90 y + 625 = 0

\frac{( x + 2 ) ^{2}}{16} + \frac{y ^{2}}{49} = 1

What is the center 3x^2+4y^2+24x-16y+52=0 ?
The center 3x^2+4y^2+24x-16y+52=0 is (-4,2)