3x^2+y^2=x+y

Lösung

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

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

Schritte zur Lösung

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

9 x^{2} + 26 y^{2} = 60

8 x^{2} + 9 y^{2} - 16 x - 54 y + 17 = 0

\frac{( x + 3 ) ^{2}}{21} + \frac{( y - 2 ) ^{2}}{49} = 1

16 - x^{2} - 9 y^{2} \geq 0

9 x^{2} + 6 y^{2} + 54 x - 36 y + 81 = 0