What is the center 25x^2+4y^2-50x+16y-59=0 ?

The center 25x^2+4y^2-50x+16y-59=0 is (1,-2)

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center 25x^2+4y^2-50x+16y-59=0

center

25 x^{2} + 4 y^{2} - 50 x + 16 y - 59 = 0

(1, - 2)

Solution steps

25 x^{2} + 4 y^{2} - 50 x + 16 y - 59 = 0

Rewrite

25 x^{2} + 4 y^{2} - 50 x + 16 y - 59 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

And the center is:

(1, - 2)

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

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

9 x^{2} + 16 y^{2} = 576

\frac{( x - 10 ) ^{2}}{4} + \frac{y ^{2}}{16} = 1

12 x^{2} + 8 x + 8 y^{2} + 12 y = 0

What is the center 25x^2+4y^2-50x+16y-59=0 ?
The center 25x^2+4y^2-50x+16y-59=0 is (1,-2)