What is the center 12x^2+20y^2-12x+40y-37=0 ?

The center 12x^2+20y^2-12x+40y-37=0 is (1/2 ,-1)

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

center

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

(\frac{1}{2}, - 1)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (\frac{1}{2}, - 1), a = 5, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (\frac{1}{2}, - 1), a = 5, b = 3

And the center is:

(\frac{1}{2}, - 1)

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

x^{2} + y^{2} - 8 y = 0

\frac{( x - 1 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{5} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1

What is the center 12x^2+20y^2-12x+40y-37=0 ?
The center 12x^2+20y^2-12x+40y-37=0 is (1/2 ,-1)