x^2+(3-i)z+2-6i=0

解

x^{2} + (3 - i) z + 2 - 6 i = 0

(h, k) = (0, - \frac{6}{5} + i \frac{8}{5}), p = - \frac{3}{4} + i \frac{1}{4}

解答ステップ

x^{2} + (3 - i) z + 2 - 6 i = 0

x^{2} + (3 - i) z + 2 - 6 i = 0

を標準的な形式で書き換える:

4 (- \frac{3}{4} + i \frac{1}{4}) (z - (- \frac{6}{5} + i \frac{8}{5})) = (x - 0)^{2}

ゆえに放物線の特性は:

(h, k) = (0, - \frac{6}{5} + i \frac{8}{5}), p = - \frac{3}{4} + i \frac{1}{4}