y^2+12x-4y+1=0

Lösung

y^{2} + 12 x - 4 y + 1 = 0

(h, k) = (\frac{1}{4}, 2), p = - 3

Schritte zur Lösung

y^{2} + 12 x - 4 y + 1 = 0

Rewrite

y^{2} + 12 x - 4 y + 1 = 0

in the standard form

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

Therefore parabola properties are:

(h, k) = (\frac{1}{4}, 2), p = - 3

y^{2} = - 4 (X - 1)

x^{2} - 4 y + 6 x - 7 = 0

\frac{v}{s} = \frac{v}{s + 1} + 1

56 y = 12 x^{2}

4 y^{2} - 8 x = 0