x^2+8x-4y+12=0

Lösung

x^{2} + 8 x - 4 y + 12 = 0

(h, k) = (- 4, - 1), p = 1

Schritte zur Lösung

x^{2} + 8 x - 4 y + 12 = 0

Rewrite

x^{2} + 8 x - 4 y + 12 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (- 4, - 1), p = 1

y^{2} - 6 y + 2 x + 5 = 0

x = y^{2} - 4 y + 1

x - 1 = y^{2}

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

y \geq x^{2} + 6