(3y+3)^2+x=(x-4)^2

解

(3 y + 3)^{2} + x = (x - 4)^{2}

(h, k) = (\frac{9}{2}, - 1), a = \frac{17}{2}, b = \frac{17}{6}

解答ステップ

(3 y + 3)^{2} + x = (x - 4)^{2}

(3 y + 3)^{2} + x - (x - 4)^{2} = 0

を標準的な双曲線のequationで書き換える

\frac{( x - \frac{9}{2} ) ^{2}}{( \frac{17}{2} ) ^{2}} - \frac{( y - ( - 1 ) ) ^{2}}{( \frac{17}{6} ) ^{2}} = 1

ゆえに双曲線の特性は:

(h, k) = (\frac{9}{2}, - 1), a = \frac{17}{2}, b = \frac{17}{6}