椭圆的性质问题,速求夜等
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椭圆的性质问题,速求夜等
椭圆上一点M(x,y),焦点在x轴,MF1=a+ex,MF2=a-ex,这个是咋推理的来着,速度速度
椭圆上一点M(x,y),焦点在x轴,MF1=a+ex,MF2=a-ex,这个是咋推理的来着,速度速度
设F1(c,0)
证明:
|MF1|2
=(x - c)2 + y2
=[a2(x - c)2 + a2y2]/a2
=[a2x2 - 2a2cx + a2c2 + a2y2]/a2 /***--根据b2x2 + a2y2 = a2b2 ***/
=[a2x2 - 2a2cx + a2c2 + a2b2 - b2x2]/a2
=[(a2-b2)x2 = 2a2cx + a2(b2 + c2)]/a2
=[c2x2 -2a2cx + a^4]/a2
=(a2 - cx)2/a2
∴PF1 = (a2 - cx)/a = a - (c/a)x = a - ex
同理可证:PF2 = a + ex
证明:
|MF1|2
=(x - c)2 + y2
=[a2(x - c)2 + a2y2]/a2
=[a2x2 - 2a2cx + a2c2 + a2y2]/a2 /***--根据b2x2 + a2y2 = a2b2 ***/
=[a2x2 - 2a2cx + a2c2 + a2b2 - b2x2]/a2
=[(a2-b2)x2 = 2a2cx + a2(b2 + c2)]/a2
=[c2x2 -2a2cx + a^4]/a2
=(a2 - cx)2/a2
∴PF1 = (a2 - cx)/a = a - (c/a)x = a - ex
同理可证:PF2 = a + ex