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

(\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)

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)