2x^2+y>0

Lösung

2 x^{2} + y > 0

(h, k) = (0, 0), p = - \frac{1}{8}

Schritte zur Lösung

2 x^{2} + y > 0

Rewrite

2 x^{2} + y > 0

in the standard form:

4 (- \frac{1}{8}) (y - 0) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = - \frac{1}{8}

(x - 0) + (y - 8)^{2} = 25

y - 1 = - \frac{1}{3} x (x - 4)

(x - 4)^{2} = 12 (y + 1)

2 x^{2} - y + 7 = 0

20 x - 4 y = - 4 x^{2}