已知椭圆X²/2+Y²=1 过点A(2,1)引椭圆割线,求截得弦的中点的轨迹方程
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2013-04-28
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过A(2,1)引椭圆割线BC,求截得弦中点P的轨迹方程
P(x,y)
xB+xC=2x,yB+yC=2y
k(BC)=(yB-yC)/(xB-xC)=(yA-yP)/(xA-xP)=(1-y)/(2-x)
(xB+xC)+2(yB+yC)*(yB-yC)/(xB-xC)=0
2x+2*2y*(1-y)/(2-x)=0
(x-1)^2+2*(y-0.5)^2=1.5
弦中点的轨迹方程椭圆:(x-1)^2/1.5+(y-0.5)/0.75=1
P(x,y)
xB+xC=2x,yB+yC=2y
k(BC)=(yB-yC)/(xB-xC)=(yA-yP)/(xA-xP)=(1-y)/(2-x)
(xB+xC)+2(yB+yC)*(yB-yC)/(xB-xC)=0
2x+2*2y*(1-y)/(2-x)=0
(x-1)^2+2*(y-0.5)^2=1.5
弦中点的轨迹方程椭圆:(x-1)^2/1.5+(y-0.5)/0.75=1
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