设A1、A2是双曲线x2/4-y2=1的实轴两个端点,P1、P2是垂直于A1A2的弦的两个端点,则直线A1P1与A2P2交点的轨迹方程为( )A.x2/4+y2=1 B.x2+y2/4=1 C.x2-y2/4=1 D.x2/4-y2=1
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 22:37:12
设A1、A2是双曲线x2/4-y2=1的实轴两个端点,P1、P2是垂直于A1A2的弦的两个端点,则直线A1P1与A2P2交点的轨迹方程为( )A.x2/4+y2=1 B.x2+y2/4=1 C.x2-y2/4=1 D.x2/4-y2=1
设A1、A2是双曲线x2/4-y2=1的实轴两个端点,P1、P2是垂直于A1A2的弦的两个端点,则直线A1P1与A2P2交点的
轨迹方程为( )
A.x2/4+y2=1 B.x2+y2/4=1 C.x2-y2/4=1 D.x2/4-y2=1
设A1、A2是双曲线x2/4-y2=1的实轴两个端点,P1、P2是垂直于A1A2的弦的两个端点,则直线A1P1与A2P2交点的轨迹方程为( )A.x2/4+y2=1 B.x2+y2/4=1 C.x2-y2/4=1 D.x2/4-y2=1
A1(-2,0),A2(2,0),设P1(x0,y0),P2(x0,-y0),A1P1与A2P2交点P(x,y)
A1,P1,P三点共线:y/(x+2)=y0/(x0+2)
A2,P2,P三点共线:y/(x-2)=y0/(2-x0)
解得x0=4/x,y0=2y/x,代入双曲线方程得x^2+4y^2=4,即x^2/4+y^2=1
答案是A
设p1的坐标为(X,Y),则p2为(X,-Y),那么A1P1的斜率为Y/(X+2),A2P2的斜率为-Y/(X-2),
A1P1和A2P2的斜率之积为-(Y2)/(X2-4).而X,Y满足x2/4-y2=1。则-(Y2)/(X2-4)=-1/4
记A1P1和A2P2交点为T,则A1T和A2T的斜率之积为-1/4.
可知,T在以A1A2为顶点,-b2/a2=-1/4的椭圆上。...
全部展开
设p1的坐标为(X,Y),则p2为(X,-Y),那么A1P1的斜率为Y/(X+2),A2P2的斜率为-Y/(X-2),
A1P1和A2P2的斜率之积为-(Y2)/(X2-4).而X,Y满足x2/4-y2=1。则-(Y2)/(X2-4)=-1/4
记A1P1和A2P2交点为T,则A1T和A2T的斜率之积为-1/4.
可知,T在以A1A2为顶点,-b2/a2=-1/4的椭圆上。所以答案为A
收起