2x^2+y^2-2x<= 0

Lösung

2 x^{2} + y^{2} - 2 x \leq 0

(h, k) = (\frac{1}{2}, 0), a = \frac{1}{2}, b = \frac{1}{2}

Schritte zur Lösung

2 x^{2} + y^{2} - 2 x \leq 0

Rewrite

2 x^{2} + y^{2} - 2 x \leq 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (\frac{1}{2}, 0), a = \frac{1}{2}, b = \frac{1}{2}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (\frac{1}{2}, 0), b = \frac{1}{2}, a = \frac{1}{2}

4 x^{2} - 16 + y^{2} + 2 y = - 1

8 x^{2} + 18 y^{2} + 32 x - 36 y - 22 = 0

4 x^{2} + 2 y^{2} = 19

4 x^{2} + 16 x + 16 y^{2} - 32 y - 32 = 0

\frac{( x - 4 ) ^{2}}{5 ^{2}} + \frac{( y - 7 ) ^{2}}{3 ^{2}} = 1