4y^2=x

Lösung

4 y^{2} = x

(h, k) = (0, 0), p = \frac{1}{16}

Schritte zur Lösung

4 y^{2} = x

Rewrite

4 y^{2} = x

in the standard form:

4 \cdot \frac{1}{16} (x - 0) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = \frac{1}{16}

x = y^{2} - 7

(x - 2)^{2} = - 8 (y - 1)

16 y^{2} + 8 y - 24 x + 49 = 0

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

2 y^{2} = 8 b x