(1)y=2x^2

Lösung

(1) y = 2 x^{2}

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

Schritte zur Lösung

1 \cdot y = 2 x^{2}

Rewrite

1 \cdot y = 2 x^{2}

in the standard form:

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

Therefore parabola properties are:

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

y^{2} = x - 1

x + 4 y^{2} - 8 = 0

x - 1 = (y + 5)^{2}

7 x^{2} + y = 13

x^{2} + 20 y = 10