center (x^2)/4+(y^2)/3 =1

Lösung

center

\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1

(0, 0)

Schritte zur Lösung

\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1

Rewrite

\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{2 ^{2}} + \frac{( y - 0 ) ^{2}}{( 3 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = 2, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 2, b = 3

And the center is:

(0, 0)

ce n t er \frac{( x - 4 ) ^{2}}{4} + \frac{y ^{2}}{16} = 1

ce n t er \frac{x ^{2}}{100} + \frac{y ^{2}}{52} = 1

ce n t er 9 x^{2} + 25 y^{2} + 36 x + 150 y + 36 = 0

ce n t er 3 x^{2} + 4 y^{2} + 12 x - 8 y + 9 = 0

ce n t er x^{2} + y^{2} \leq y