y^2+12y-4x+16=0

Lösung

y^{2} + 12 y - 4 x + 16 = 0

(h, k) = (- 5, - 6), p = 1

Schritte zur Lösung

y^{2} + 12 y - 4 x + 16 = 0

Rewrite

y^{2} + 12 y - 4 x + 16 = 0

in the standard form

4 \cdot 1 \cdot (x - (- 5)) = (y - (- 6))^{2}

Therefore parabola properties are:

(h, k) = (- 5, - 6), p = 1

(y - 5)^{2} = 12 (x + 3)

y^{2} + 2 y + 12 x + 25 = 0

y^{2} = x + 81

3 x^{2} + 3 y = 0

y^{2} - 6 x = - 14 y - 49