25y^2+40x+20y+28=0

解答

25 y^{2} + 40 x + 20 y + 28 = 0

(h, k) = (- \frac{3}{5}, - \frac{2}{5}), p = - \frac{2}{5}

求解步骤

25 y^{2} + 40 x + 20 y + 28 = 0

将

25 y^{2} + 40 x + 20 y + 28 = 0

改写为标准形式

4 (- \frac{2}{5}) (x - (- \frac{3}{5})) = (y - (- \frac{2}{5}))^{2}

因此抛物线性质是:

(h, k) = (- \frac{3}{5}, - \frac{2}{5}), p = - \frac{2}{5}

x^{2} = - \frac{1}{3} y

(y + 5)^{2} = - \frac{1}{3} (x - 3)

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

(y - (- 19))^{2} = 2 (x - 1)

24 y = 8 x^{2}