3y^2-4x+12y+16=0

Lösung

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

(h, k) = (1, - 2), p = \frac{1}{3}

Schritte zur Lösung

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

Rewrite

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

in the standard form

4 \cdot \frac{1}{3} (x - 1) = (y - (- 2))^{2}

Therefore parabola properties are:

(h, k) = (1, - 2), p = \frac{1}{3}