5y^2+2x-21y+20=0

解

5 y^{2} + 2 x - 21 y + 20 = 0

(h, k) = (\frac{41}{40}, \frac{21}{10}), p = - \frac{1}{10}

解答ステップ

5 y^{2} + 2 x - 21 y + 20 = 0

5 y^{2} + 2 x - 21 y + 20 = 0

を標準的な形式で書き換える

4 (- \frac{1}{10}) (x - \frac{41}{40}) = (y - \frac{21}{10})^{2}

ゆえに放物線の特性は:

(h, k) = (\frac{41}{40}, \frac{21}{10}), p = - \frac{1}{10}