leitkurve y^2=5y+7x+3

Lösung

leitkurve

y^{2} = 5 y + 7 x + 3

x = - \frac{43}{14}

Schritte zur Lösung

Rewrite

y^{2} = 5 y + 7 x + 3

in the standard form

4 \cdot \frac{7}{4} (x - (- \frac{37}{28})) = (y - \frac{5}{2})^{2}

(h, k) = (- \frac{37}{28}, \frac{5}{2}), p = \frac{7}{4}

Parabola is symmetric around the x-axis and so the directrix is a line parallel to the y-axis, a distance

- p

from the center

(- \frac{37}{28}, \frac{5}{2})

x-coordinate

x = - \frac{37}{28} - p

x = - \frac{37}{28} - \frac{7}{4}

Fasse zusammen

x = - \frac{43}{14}

d i rec t r i x 2 y^{2} - 12 y - x + 5 = 0

d i rec t r i x - 0.03 x^{2} + 45 x + 124.5

d i rec t r i x 4 y^{2} - 12 y + 56 x + 177 = 0

d i rec t r i x 3 x^{2} + 19 x - 40

d i rec t r i x (y - 10)^{2} = - 12 (x + 2)