已知A,B为两定点,动点M到A与到B的距离比为常数λ,求点M的轨迹方程,并注明轨迹是什么曲线
2个回答
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M(x,y) A(-a,0) B(a,0) a>0
AM/BM=λ
(AM/BM)^2=λ^2
AM^2=(λBM)^2
{(x+a)^2+y^2]=[(x-a)^2+y^2]λ^2
(λ^2-1)x^2-(2aλ^2+2a)x+(λ^2-1)a^2+(λ^2-1)y^2=0
(λ^2-1)[x-a(λ^2+1)/(λ^2-1)]^2-a^2(λ^2+1)^2/(λ^2-1)+(λ^2-1)a^2+(λ^2-1)y^2=0
其中:-a^2(λ^2+1)^2/(λ^2-1)+(λ^2-1)a^2
=a^2[(λ^2-1)^2/(λ^2-1)-(λ^2+1)^2/(λ^2-1)]
=a^2[(λ^2-1-λ^2-1)(λ^2-1+λ^2+1)/(λ^2-1)]
=a^2(-2)(2λ^2)/(λ^2-1)
=-4(aλ)^2/(λ^2-1)
(λ^2-1)[x-a(λ^2+1)/(λ^2-1)]^2+(λ^2-1)y^2=4(aλ)^2/(λ^2-1)
[x-a(λ^2+1)/(λ^2-1)]^2+y^2=4(aλ)^2/(λ^2-1)^2
这是以(a(λ^2+1)/(λ^2-1),0)为圆心,2aλ/(λ^2-1)为半径的圆。λ不等于1
当λ=1时:就AB的垂直平分线,即x=0
AM/BM=λ
(AM/BM)^2=λ^2
AM^2=(λBM)^2
{(x+a)^2+y^2]=[(x-a)^2+y^2]λ^2
(λ^2-1)x^2-(2aλ^2+2a)x+(λ^2-1)a^2+(λ^2-1)y^2=0
(λ^2-1)[x-a(λ^2+1)/(λ^2-1)]^2-a^2(λ^2+1)^2/(λ^2-1)+(λ^2-1)a^2+(λ^2-1)y^2=0
其中:-a^2(λ^2+1)^2/(λ^2-1)+(λ^2-1)a^2
=a^2[(λ^2-1)^2/(λ^2-1)-(λ^2+1)^2/(λ^2-1)]
=a^2[(λ^2-1-λ^2-1)(λ^2-1+λ^2+1)/(λ^2-1)]
=a^2(-2)(2λ^2)/(λ^2-1)
=-4(aλ)^2/(λ^2-1)
(λ^2-1)[x-a(λ^2+1)/(λ^2-1)]^2+(λ^2-1)y^2=4(aλ)^2/(λ^2-1)
[x-a(λ^2+1)/(λ^2-1)]^2+y^2=4(aλ)^2/(λ^2-1)^2
这是以(a(λ^2+1)/(λ^2-1),0)为圆心,2aλ/(λ^2-1)为半径的圆。λ不等于1
当λ=1时:就AB的垂直平分线,即x=0
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