直线交椭圆于AB OA垂直OB 点C的轨迹方程
1个回答
A(x1,y1)B(x2,y2)
y=kx+m
b^2x^2+a^2y^2=a^2b^2
b^2x^2+a^2(kx+m)^2=a^2b^2
(b^2+a^2k^2)x^2+2kma^2x+a^2m^2-a^2b^2=0
x1+x2=-2kma^2/(b^2+a^2k^2),x1x2=a^2(m^2-b^2)/(b^2+a^2k^2)
koa=y1/x1,kob=y2/x2
koakob=-1
y1/x1*y2/x2=-1
(kx1+m)(kx2+m)/x1x2=-1
(k^2x1x2+m^2+mk(x1+x2))/x1x2=-1
k^2+m^2/x1x2+mk(x1+x2)/x1x2=-1
k^2+m^2(b^2+a^2k^2)/a^2(m^2-b^2)+mk*(-2km/(m^2-b^2)=-1
a^2k^2b^2-a^2m^2+a^2b^2-m^2b^2=0(1)
koc=y/x=-1/k
k=-y/x
y=kx+m
m=2y
k=-x/y
代入（1）
a^2b^2(x/y)^2-a^24y^2+a^2b^2-4y^2b^2=0
a^2b^2x^2-4a^2y^4+a^2b^2y^2-4b^2y^4=0
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