What is the center 25x^2+16y^2+50x-64y-311=0 ?

The center 25x^2+16y^2+50x-64y-311=0 is (-1,2)

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center 25x^2+16y^2+50x-64y-311=0

center

25 x^{2} + 16 y^{2} + 50 x - 64 y - 311 = 0

(- 1, 2)

Solution steps

25 x^{2} + 16 y^{2} + 50 x - 64 y - 311 = 0

Rewrite

25 x^{2} + 16 y^{2} + 50 x - 64 y - 311 = 0

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{4 ^{2}} + \frac{( y - 2 ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

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

And the center is:

(- 1, 2)

4 x^{2} + 9 y^{2} - 16 x - 18 y - 11 = 0

\frac{( x - 7 ) ^{2}}{25} + \frac{( y - 6 ) ^{2}}{39} = 1

\frac{( x - 1 ) ^{2}}{100} + \frac{( y - 2 ) ^{2}}{36} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{64} = 1

4 x^{2} + 5 y^{2} - 2 x + 78 y - 83 = x - 3 y + 3

What is the center 25x^2+16y^2+50x-64y-311=0 ?
The center 25x^2+16y^2+50x-64y-311=0 is (-1,2)