What is the center 4y^2+2x^2-4y-8x-15=0 ?

The center 4y^2+2x^2-4y-8x-15=0 is (2, 1/2)

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center 4y^2+2x^2-4y-8x-15=0

center

4 y^{2} + 2 x^{2} - 4 y - 8 x - 15 = 0

(2, \frac{1}{2})

Solution steps

4 y^{2} + 2 x^{2} - 4 y - 8 x - 15 = 0

Rewrite

4 y^{2} + 2 x^{2} - 4 y - 8 x - 15 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, \frac{1}{2}), a = 23, b = 6

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, \frac{1}{2}), a = 23, b = 6

And the center is:

(2, \frac{1}{2})

\frac{( x - 2 ) ^{2}}{16} + \frac{y ^{2}}{9} = 1

\frac{( x - 3 ) ^{2}}{4} + \frac{( y - 2 ) ^{2}}{9} = 1

36 x^{2} + 9 y^{2} + 48 x - 36 y + 43 = 0

\frac{( x - 3 ) ^{2}}{4} + \frac{( y - 2 ) ^{2}}{9} = 3

5 x^{2} + 3 y^{2} + 30 x + 60 y - 45 = 0

What is the center 4y^2+2x^2-4y-8x-15=0 ?
The center 4y^2+2x^2-4y-8x-15=0 is (2, 1/2)