y^2+4y-x+4=0

Lösung

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

(h, k) = (0, - 2), p = \frac{1}{4}

Schritte zur Lösung

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

Rewrite

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

in the standard form

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

Therefore parabola properties are:

(h, k) = (0, - 2), p = \frac{1}{4}

x^{2} + 6 x - 4 y - 3 = 0

x^{2} - 3 y = 0

x^{2} - 3 y = 4

(x - 5)^{2} = - 20 (y - 1)

y = \frac{4 x + 2}{y}