2x^2=7y-4

Lösung

2 x^{2} = 7 y - 4

(h, k) = (0, \frac{4}{7}), p = \frac{7}{8}

Schritte zur Lösung

2 x^{2} = 7 y - 4

Rewrite

2 x^{2} = 7 y - 4

in the standard form:

4 \cdot \frac{7}{8} (y - \frac{4}{7}) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, \frac{4}{7}), p = \frac{7}{8}

2 x^{2} + 4 x + z = 0

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

y = 3 x^{2} - 30 x + 68

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

x = - 3 y^{2} + 6 y + 2