4t=1+15r^2

Lösung

4 t = 1 + 15 r^{2}

(h, k) = (\frac{1}{4}, 0), p = \frac{1}{15}

Schritte zur Lösung

4 t = 1 + 15 r^{2}

Rewrite

4 t = 1 + 15 r^{2}

in the standard form:

4 \cdot \frac{1}{15} (t - \frac{1}{4}) = (r - 0)^{2}

Therefore parabola properties are:

(h, k) = (\frac{1}{4}, 0), p = \frac{1}{15}

33 x - 2 y^{2} = y^{2} + 18 x

2 x^{2} + 4 x + 5 k (x) = 0

x^{2} + 2 x + 4 y - 19 = 0

x + y^{2} = 6

y \leq x^{2} - 3