2v=3u^2

Lösung

2 v = 3 u^{2}

(h, k) = (0, 0), p = \frac{1}{6}

Schritte zur Lösung

2 v = 3 u^{2}

Rewrite

2 v = 3 u^{2}

in the standard form:

4 \cdot \frac{1}{6} (v - 0) = (u - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = \frac{1}{6}

4 x - 11 y^{2} = 0

x^{2} = 8 (y + 2)

x - y = - 3 x^{2}

y + 3 x + 5 = - 6 x + 6 x^{2}

x^{(2)} + 8 x + 16 y - 16 = 0