题目内容
已知点A(0,-2),B(0,4),动点P(x,y)满足
•
=y2-8.
(1)求动点P的轨迹方程;
(2)设(1)中所求轨迹与直线y=x+b交于C、D两点,且OC⊥OD(O为原点),求b的值.
PA |
PB |
(1)求动点P的轨迹方程;
(2)设(1)中所求轨迹与直线y=x+b交于C、D两点,且OC⊥OD(O为原点),求b的值.
(1)∵动点P(x,y)满足
•
=y2-8,
∴(-x,-2-y)•(-x,4-y)=y2-8,
∴x2+y2-2y-8=y2-8,化为x2=2y.
∴动点P的轨迹方程为x2=2y;
(2)设C(x1,y1),D(x2,y2).联立
,
化为x2-2x-2b=0,∴△=4+8b>0.
∴x1+x2=2,x1x2=-2b.(*)
∵
⊥
,∴x1x2+y1y2=x1x2+(x1+b)(x2+b)=0,
化为2x1x2+b(x1+x2)+b2=0,
把(*)代入上式得-4b+2b+b2=0,解得b=0或2.满足△>0.
∴b=0或2.
PA |
PB |
∴(-x,-2-y)•(-x,4-y)=y2-8,
∴x2+y2-2y-8=y2-8,化为x2=2y.
∴动点P的轨迹方程为x2=2y;
(2)设C(x1,y1),D(x2,y2).联立
|
化为x2-2x-2b=0,∴△=4+8b>0.
∴x1+x2=2,x1x2=-2b.(*)
∵
OC |
OD |
化为2x1x2+b(x1+x2)+b2=0,
把(*)代入上式得-4b+2b+b2=0,解得b=0或2.满足△>0.
∴b=0或2.
练习册系列答案
相关题目