4y^2-5x-12=0

Lösung

4 y^{2} - 5 x - 12 = 0

(h, k) = (- \frac{12}{5}, 0), p = \frac{5}{16}

Schritte zur Lösung

4 y^{2} - 5 x - 12 = 0

Rewrite

4 y^{2} - 5 x - 12 = 0

in the standard form:

4 \cdot \frac{5}{16} (x - (- \frac{12}{5})) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (- \frac{12}{5}, 0), p = \frac{5}{16}

3 x^{2} + y = 5

a = \frac{2 x}{t ^{2}}

y^{(2)} - 10 y - 5 x + 15 = 0

x^{2} + y = - 1

5 x^{2} + 30 x + 24 y = 51