maths~~~~~~~~~~~~~~~~~
2010-10-06 3:19 am
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回答 (2)
2010-10-06 3:38 am
✔ 最佳答案
設 M 為 AB 中點 :OA^2 = x^2 + y^2 .....(1);PA^2 = x^2 + PM^2 .....(2);OP^2 = PA^2 + OA^2 , 由(1) , (2) :OP^2 = (x^2 + PM^2) + (x^2 + y^2)(PM + y)^2 = (x^2 + PM^2) + (x^2 + y^2)PM^2 + 2y * PM + y^2 = 2x^2 + PM^2 + y^22y * PM = 2x^2PM = (x^2)/y 
2010-10-06 7:24 am
Let M be the mid-point of AB.
Then triangle OMA is similar to triangle OAP.
Thus we have OA^2 = OM*OP
so OP = OA^2 / OM = [x^2+y^2]/y.
Then triangle OMA is similar to triangle OAP.
Thus we have OA^2 = OM*OP
so OP = OA^2 / OM = [x^2+y^2]/y.
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原文連結 [永久失效]:
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