What is the center x^2+5y^2-4x+20y=-19 ?

The center x^2+5y^2-4x+20y=-19 is (2,-2)

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center x^2+5y^2-4x+20y=-19

center

x^{2} + 5 y^{2} - 4 x + 20 y = - 19

(2, - 2)

Solution steps

x^{2} + 5 y^{2} - 4 x + 20 y = - 19

Rewrite

x^{2} + 5 y^{2} - 4 x + 20 y = - 19

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, - 2), a = 5, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, - 2), a = 5, b = 1

And the center is:

(2, - 2)

\frac{( x + 4 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{25} = 1

16 x^{2} + 9 y^{2} - 64 x - 54 y + 1 = 0

3 x^{2} + 2 y^{2} - 12 x + 8 y + 19 = 0

25 x^{2} + y^{2} - 100 x + 8 y + 91 = 0

49 x^{2} + 625 y^{2} = 30625

What is the center x^2+5y^2-4x+20y=-19 ?
The center x^2+5y^2-4x+20y=-19 is (2,-2)