Question

高中数学题目`在线求解

20：已知：三角形ABC为直角三角形，角C为直角，A(0,-8),顶点C在x轴上运动，M在y轴上，向量AM=1/2(向量AB+向量AC），设B的运动轨迹为曲线E．
1）求B的运动轨迹曲线E的方程
2）过点P(2，4）的直线l与曲线E相交于不同的两点Q、N,且满足向量QP=向量PN，求直线l的方程。

Answer 1

(1)
A(0,-8) , C(c,0), M(0,m), ∠C=90°, B(x,y)

AM = (1/2)(AB+AC)
OM-OA =(1/2)(OB-OA +OC -OA)
(0,m+8) = (1/2)( (x,y) + (c, 16))
= (1/2)(x+c,y+16)
=> (1/2)(x+c)=0
c = -x
∠C=90°
=> AC.CB =0
(c,8).(x-c,y) =0
c(x-c) - 8y = 0
-x(2x) -8y =0
x^2 + 4y =0
E的方程: x^2 + 4y =0 (1)
(2)
let Q(x1,y1),N(x2,y2)
QP=PN
OP-OQ = ON-OP
(2-x1,4-y1)= (x2-2,y2-4)
=> 2-x1= x2-2 and 4-y1 = y2-4
=> x1+x2 = 4 and y1+y2 = 8
l: y= mx + c
过点P(2，4）
=> 4 = 2m +c
c = 4-2m
l: y = mx+ 4-2m (2)
sub (2) into (1)
x^2 + 4y =0
x^2 + 4(mx+ 4-2m) =0
x^2 + 4mx + 16-8m =0
x1+x2 = -4m = 4
=> m = -1
l: y = mx+ 4-2m
y = -x + 6