一条变动的直线L与椭圆X^2/4+Y^2/2=1交于P,Q两点,M是L上的动点,满足关系|MP|×|MQ|=2.若直线L在变动过程中始终保持其斜率等于1,求动点M的轨迹方程
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![一条变动的直线L与椭圆X^2/4+Y^2/2=1交于P,Q两点,M是L上的动点,满足关系|MP|×|MQ|=2.若直线L在变动过程中始终保持其斜率等于1,求动点M的轨迹方程](/uploads/image/z/6605423-71-3.jpg?t=%E4%B8%80%E6%9D%A1%E5%8F%98%E5%8A%A8%E7%9A%84%E7%9B%B4%E7%BA%BFL%E4%B8%8E%E6%A4%AD%E5%9C%86X%EF%BC%BE2%EF%BC%8F4%EF%BC%8BY%EF%BC%BE2%EF%BC%8F2%EF%BC%9D1%E4%BA%A4%E4%BA%8EP%2CQ%E4%B8%A4%E7%82%B9%2CM%E6%98%AFL%E4%B8%8A%E7%9A%84%E5%8A%A8%E7%82%B9%2C%E6%BB%A1%E8%B6%B3%E5%85%B3%E7%B3%BB%EF%BD%9CMP%EF%BD%9C%C3%97%EF%BD%9CMQ%EF%BD%9C%EF%BC%9D2%EF%BC%8E%E8%8B%A5%E7%9B%B4%E7%BA%BFL%E5%9C%A8%E5%8F%98%E5%8A%A8%E8%BF%87%E7%A8%8B%E4%B8%AD%E5%A7%8B%E7%BB%88%E4%BF%9D%E6%8C%81%E5%85%B6%E6%96%9C%E7%8E%87%E7%AD%89%E4%BA%8E1%2C%E6%B1%82%E5%8A%A8%E7%82%B9M%E7%9A%84%E8%BD%A8%E8%BF%B9%E6%96%B9%E7%A8%8B)
一条变动的直线L与椭圆X^2/4+Y^2/2=1交于P,Q两点,M是L上的动点,满足关系|MP|×|MQ|=2.若直线L在变动过程中始终保持其斜率等于1,求动点M的轨迹方程
一条变动的直线L与椭圆X^2/4+Y^2/2=1交于P,Q两点,M是L上的动点,满足关系|MP|×|MQ|=2.
若直线L在变动过程中始终保持其斜率等于1,求动点M的轨迹方程
一条变动的直线L与椭圆X^2/4+Y^2/2=1交于P,Q两点,M是L上的动点,满足关系|MP|×|MQ|=2.若直线L在变动过程中始终保持其斜率等于1,求动点M的轨迹方程
直线L:y=x+k,P(x1,y1),Q(x2,y2),M(m,n)
==> y1=x1+k,y2=x2+k,k=n-m ...(1)
y=x+k 代入X^2/4+Y^2/2=1,得:
3x^2 +4kx +(2k^2 -4) =0 ...(2)
|MP|*|MQ|=2
|MP| =根号[(x1-m)^2+(y1-n)^2] =根号[2*(x1-m)^2]
|MQ| =根号[(x2-m)^2+(y2-n)^2] =根号[2*(x2-m)^2]
1 =|(x1-m)(x2-m)| =|x1x2 -(x1+x2)m +m^2| ...(3)
(1)(2)(3) ==>
m^2 +2*n^2 = 1,or 7
因此,动点M的轨迹为椭圆:
x^2 +2*y^2 =1,及:x^2 +2*y^2 =7
直线L:y=x k, P(x1,y1), Q(x2,y2), M(m,n) ==> y1=x1 k, y2=x2 k, k=n-m ...(1) y=x k 代入X^2/4 Y^2/2=1,得: 3x^2 4kx (2k^2 -4) =0 ...(2) MP|*|MQ|=2 MP| =根号[(x1-m)^2 (y1-n)^2] =根号[2*(x1-m)^2] MQ| =根号[(x2-m)^2 ...
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直线L:y=x k, P(x1,y1), Q(x2,y2), M(m,n) ==> y1=x1 k, y2=x2 k, k=n-m ...(1) y=x k 代入X^2/4 Y^2/2=1,得: 3x^2 4kx (2k^2 -4) =0 ...(2) MP|*|MQ|=2 MP| =根号[(x1-m)^2 (y1-n)^2] =根号[2*(x1-m)^2] MQ| =根号[(x2-m)^2 (y2-n)^2] =根号[2*(x2-m)^2] 1 =|(x1-m)(x2-m)| =|x1x2 -(x1 x2)m m^2| ...(3)(1)(2)(3) ==> m^2 2*n^2 = 1,or 7 因此,动点M的轨迹为椭圆 x^2 2*y^2 =1, 及:x^2 2*y^2 =7
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