What is the center x^2+7x+3y^2+y=0 ?

The center x^2+7x+3y^2+y=0 is (-7/2 ,-1/6)

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center x^2+7x+3y^2+y=0

center

x^{2} + 7 x + 3 y^{2} + y = 0

(- \frac{7}{2}, - \frac{1}{6})

Solution steps

x^{2} + 7 x + 3 y^{2} + y = 0

Rewrite

x^{2} + 7 x + 3 y^{2} + y = 0

in the form of the standard ellipse equation

\frac{( x - ( - \frac{7}{2} ) ) ^{2}}{( \frac{37}{3} ) ^{2}} + \frac{( y - ( - \frac{1}{6} ) ) ^{2}}{( \frac{37}{3} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- \frac{7}{2}, - \frac{1}{6}), a = \frac{37}{3}, b = \frac{37}{3}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- \frac{7}{2}, - \frac{1}{6}), a = \frac{37}{3}, b = \frac{37}{3}

And the center is:

(- \frac{7}{2}, - \frac{1}{6})

9 x^{2} + 4 y^{2} = 36

9 X^{2} + 4 Y^{2} = 36

41 x^{2} + 16 y^{2} + 246 x - 192 y + 289 = 0

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

25 x^{2} + 16 y^{2} = 400