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

因此抛物线性质是:

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

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

9 L (x) = \frac{x}{3 L ( x )}

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

3 x^{2} - y + 6 = 0

x^{2} - 2 x + 4 y - 11 = 0