x+y=y^2

Lösung

x + y = y^{2}

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

Schritte zur Lösung

x + y = y^{2}

Rewrite

x + y = y^{2}

in the standard form

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

Therefore parabola properties are:

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

x^{2} + 6 x - 2 y + 15 = 0

y \leq - x^{2} - 5 x - \frac{13}{4}

x^{2} + 3 x + 5 y - 5 = 0

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

y = - (x - 2)^{2} - 6