我根据椭圆是到固定两点之和等于常数的连续轨迹得出了椭圆方程

送交者: HunHunSheng 于 2009-03-12, 14:24:07:

sqrt((x-e)^2+y^2)=2*a-sqrt((x+e)^2+y^2).

where e
Eventually I got

x^2/a^2+Y^2/b^2=1

但我还是事先画了一个左右上下对称的椭圆根据
对称先算出常数等于两倍的半长轴，根据勾股定理
得出，纯粹从椭圆是到固定两点之和等于常数的连续轨迹的
定义我能事先做这些假设么？

你直接把那个式子平方两次不就得到椭圆方程了吗？为什么要画图呢？ (无内容) - 008 (0 bytes) 2009-03-12, 15:17:08 (305869)
我曾经利用几何原理，在岸边推断出一个近海油架的距离，断定可以划KAYAK到达 - meiyou (12 bytes) 2009-03-12, 14:34:53 (305825)
e is half-distance between two foci - HunHunSheng (277 bytes) 2009-03-12, 14:27:53 (305817)
- e^2=a^2-b^2 is not an asumption, it is derived from 椭圆是到固定两点之和 - chouqilozi (43 bytes) 2009-03-12, 14:38:45 (305828)
  - How do you derive it? (无内容) - HunHunSheng (0 bytes) 2009-03-12, 14:44:58 (305831)
    - here - chouqilozi (357 bytes) 2009-03-12, 14:53:19 (305841)
      - the triangle is messed up. when the point is in the middle of the two focal - chouqilozi (64 bytes) 2009-03-12, 14:56:14 (305847)
      - 最后两行假设了对称性。可是你凭什么可以假设对称？ (无内容) - HunHunSheng (0 bytes) 2009-03-12, 14:55:35 (305846)
        
        Didn't assume, because - chouqilozi (108 bytes) 2009-03-12, 14:58:15 (305851)
        
        那你怎么知道轨迹会和两个焦点在一条直线上？ (无内容) - HunHunSheng (0 bytes) 2009-03-12, 15:12:48 (305860)
        
        昏君you lost me. (无内容) - chouqilozi (0 bytes) 2009-03-12, 15:18:59 (305874)