4x-x^2+2y=0

Lösung

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

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

Schritte zur Lösung

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

Rewrite

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

in the standard form

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

Therefore parabola properties are:

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

x^{2} + 5 y = 20

x^{2} = x + y

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

x^{2} + 5 y = 10

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