彷徨离婚路上
幼苗
共回答了20个问题采纳率:95% 举报
直线l与x轴交点M,∵PM⊥F1F2,PF1⊥PF2,∴2*F1F2*PM = S△PF1F2 = 2*PF1*PF2
∴2*2c * PM = 2*4ab,∴PM = 2ab/c,
∴k(PF1) * k(PF2) = -1 = {(2ab/c)/[(a^2/c) + c]} * {(2ab/c)/[(a^2/c) - c]}
∴c^2 - (a^4/c^2) = 4(ab)^2/c^2,∴c^4 - a^4 = 4a^2b^2 = 4a^2 * (c^2 - a^2)
∴c^4 - 4a^2c^2 + 3a^4 = 0,结合c>a取正数解c^2 = 3a^2,∴e = √3
1年前
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