What is the center 4x^2+y^2+12x+6y+5=0 ?

The center 4x^2+y^2+12x+6y+5=0 is (-3/2 ,-3)

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center 4x^2+y^2+12x+6y+5=0

center

4 x^{2} + y^{2} + 12 x + 6 y + 5 = 0

(- \frac{3}{2}, - 3)

Solution steps

4 x^{2} + y^{2} + 12 x + 6 y + 5 = 0

Rewrite

4 x^{2} + y^{2} + 12 x + 6 y + 5 = 0

in the form of the standard ellipse equation

\frac{( x - ( - \frac{3}{2} ) ) ^{2}}{( \frac{13}{2} ) ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{( 13 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- \frac{3}{2}, - 3), a = \frac{13}{2}, b = 13

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- \frac{3}{2}, - 3), b = 13, a = \frac{13}{2}

And the center is:

(- \frac{3}{2}, - 3)

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

\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

What is the center 4x^2+y^2+12x+6y+5=0 ?
The center 4x^2+y^2+12x+6y+5=0 is (-3/2 ,-3)