y^2-3x+4y+15=0

Lösung

y^{2} - 3 x + 4 y + 15 = 0

(h, k) = (\frac{11}{3}, - 2), p = \frac{3}{4}

Schritte zur Lösung

y^{2} - 3 x + 4 y + 15 = 0

Rewrite

y^{2} - 3 x + 4 y + 15 = 0

in the standard form

4 \cdot \frac{3}{4} (x - \frac{11}{3}) = (y - (- 2))^{2}

Therefore parabola properties are:

(h, k) = (\frac{11}{3}, - 2), p = \frac{3}{4}

x = - 3 y^{2} + 8 y - 13

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

x + \frac{1}{5} y^{2} = 0

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

y = 3 x^{2} - 4 y