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

Lösung

center

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

(0, - 1)

Schritte zur Lösung

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{2 ^{2}} + \frac{( y - ( - 1 ) ) ^{2}}{1 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, - 1), a = 2, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, - 1), a = 2, b = 1

And the center is:

(0, - 1)

ce n t er \frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1

ce n t er \frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1

ce n t er 9 x^{2} + 4 y^{2} + 54 x - 8 y + 49 = 0

ce n t er \frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{36} = 1

ce n t er 16 x^{2} + 25 y^{2} - 128 x + 150 y + 81 = 0