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直线l经过点A(1,2),交椭圆X^2/36+Y^2/16=1交于点P1,P2,求P1,P2中点的轨迹方程
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直线l经过点A(1,2),交椭圆X^2/36+Y^2/16=1交于点P1,P2,求P1,P2中点的轨迹方程
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答案和解析
M(x,y)
x1+x2=2x,y1=y2=2y
k(P1P2)=k(AM)
(y1-y2)/(x1-x2)=(y-2)/(x-1)
[(x1)^2/36+(y1)^2/16)]-[(x2)^2/36+(y2)^2/16]=1-1
16(x1+x2)+36*(y1+y2)*(y1-y2)/(x1-x2)=0
16*2x+36*2y*(y-2)/(x-1)=0
(x-0.5)^2/(10/4)+(y-1)^2/(10/9)=1
x1+x2=2x,y1=y2=2y
k(P1P2)=k(AM)
(y1-y2)/(x1-x2)=(y-2)/(x-1)
[(x1)^2/36+(y1)^2/16)]-[(x2)^2/36+(y2)^2/16]=1-1
16(x1+x2)+36*(y1+y2)*(y1-y2)/(x1-x2)=0
16*2x+36*2y*(y-2)/(x-1)=0
(x-0.5)^2/(10/4)+(y-1)^2/(10/9)=1
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