What is the center 16x^2+4y^2-32x+24y-12=0 ?

The center 16x^2+4y^2-32x+24y-12=0 is (1,-3)

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center 16x^2+4y^2-32x+24y-12=0

center

16 x^{2} + 4 y^{2} - 32 x + 24 y - 12 = 0

(1, - 3)

Solution steps

16 x^{2} + 4 y^{2} - 32 x + 24 y - 12 = 0

Rewrite

16 x^{2} + 4 y^{2} - 32 x + 24 y - 12 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

And the center is:

(1, - 3)

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

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

x^{2} + 3 y^{2} = 1

25 x^{2} + 16 y^{2} = 400

x^{2} + y^{2} - 10 y = 0

What is the center 16x^2+4y^2-32x+24y-12=0 ?
The center 16x^2+4y^2-32x+24y-12=0 is (1,-3)