一道高二数学题（求点的轨迹方程）
1个回答
直线L过定点A（0,3)
曲线C:y^2= 4x
Let P(x,y),P(x1,y1),P2(x2,y2)
P is mid point of P1P2,=>
x1+x2 = 2x,y1+y2=2y
P1,P2 is on C,then
y1^2=4x1 and y2^2=4x2
=> 4(x2-x1)= y2^2-y1^2
=> (y2-y1)/(x2-x1) = 4/(y2+y1)
=> (y2-y1)/(x2-x1) = 4/2y
slope of AP = (y-3)/x
slope of P1P2 = (y2-y1)/(x2-x1)
AP 垂至于P1P2 => slope of AP x slope of P!P2 = -1 =>
((y-3)/x) ( y2-y1)/(x2-x1) = -1
((y-3)/x) (4/2y) = -1
2(y-3) = -xy
xy-2y-6 = 0
求直线L与动弦p1p2的交点M的轨迹方程:
xy-2y-6 = 0
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