在平面直角坐标系xoyo中双曲线c1;x^2-y^2=1.设斜率为2的直线l交于c1于p.q两点,若l与圆x^2+y^2
2个回答
（1）设l:2x-y+m=0,它与圆x^2+y^2=5相切,
∴|m|/√5=√5,m=土5.
把y=2x+5代入x^-y^=1,得3x^+20x-24=0,
设P(x1,y1),Q(x2,y2),则
x1+x2=-20/3,x1x2=-8,
OP⊥OQ,
(2)C2:3x^2+y^2=1
x^2/(1/3)+y^2=1
M(x1,y1) ;N(x2,y2)
直线OM: y=kx 代入 x^2-y^2=1; x1^2 = 1/(1-k^2),因此 y2^2 = k^2/(1-k^2)
直线ON: y=-1/kx,
x=-ky 代入 3x^2+y^2 = 1; (3k^2+1)y2^2=1; y2^2 = 1/(3k^2+1); x2^2 = k^2/(3k^2+1)
OM^2 = x1^2+y1^2 = (1+k^2)/(1-k^2)
ON^2 = x2^2+y2^2 = (1+k^2)/(3k^2+1)
MN^2 = OM^2+ON^2 = (1+k^2) * [3k^2+1+1-k^2]/(1-k^2) (3k^2+1) = 2(1+k^2)^2/[(1-k^2) (3k^2+1)]
OM^2*ON^2 = (1+k^2)^2/[(1-k^2) (3k^2+1)]
所以 d^2 = OM^2*ON^2/MN^2 = 1/2
d=√2/2为定值
很高兴为您解答,祝你学习进步!【学习宝典】团队为您答题.
有不明白的可以追问!如果您认可我的回答.
请点击下面的【选为满意回答】按钮,谢谢!
相关问题
