f(x)(b-1)=x^2+7

解答

f (x) (b - 1) = x^{2} + 7

(h, k) = (0, \frac{7}{b - 1}), p = \frac{b - 1}{4}

求解步骤

y (b - 1) = x^{2} + 7

将

y (b - 1) = x^{2} + 7

改写为标准形式:

4 \cdot \frac{b - 1}{4} (y - \frac{7}{b - 1}) = (x - 0)^{2}

因此抛物线性质是:

(h, k) = (0, \frac{7}{b - 1}), p = \frac{b - 1}{4}

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

\frac{1}{t + 5} = \frac{6}{r ^{2} - 25}

y^{2} - 8 y + 8 x + 16 = 0

25 = \frac{x}{y ^{2}}

x^{2} - 76 x + 20 y + 50 = 0