x^2= 125/4 (y-20)

解答

x^{2} = \frac{125}{4} (y - 20)

(h, k) = (0, 20), p = \frac{125}{16}

求解步骤

x^{2} = \frac{125}{4} (y - 20)

将

x^{2} = \frac{125}{4} (y - 20)

改写为标准形式:

4 \cdot \frac{125}{16} (y - 20) = (x - 0)^{2}

因此抛物线性质是:

(h, k) = (0, 20), p = \frac{125}{16}

\frac{x}{2} x + 5 y = 8

2 f (x) = 5 (x)^{2} - (x) + 4

x + 2 (y - 1)^{2} = 0

x^{2} = - \frac{125}{4} (y - 20)

(y - 4)^{2} = 4 - 4 x