Question

如圖所示,拋物線C₁:x²=4y,C₂:x²=-2py(p>0).點(diǎn)M(x₀,y₀)在拋物線C₂上,過(guò)M作C₁的切線,切點(diǎn)為A,B(M為原點(diǎn)O時(shí),A,B重合于O).當(dāng)x₀=1-時(shí),切線MA的斜率為-.

(1)求p的值;
(2)當(dāng)M在C₂上運(yùn)動(dòng)時(shí),求線段AB中點(diǎn)N的軌跡方程(A,B重合于O時(shí),中點(diǎn)為O).

Answer 1

(1)2   (2) x²=y

解:(1)因?yàn)閽佄锞�C₁:x²=4y上任意一點(diǎn)(x,y)的切線斜率為y′=,且切線MA的斜率為-,
所以A點(diǎn)坐標(biāo)為.
故切線MA的方程為y=-(x+1)+ .
因?yàn)辄c(diǎn)M(1-y₀)在切線MA及拋物線C₂上,于是
y₀=-(2-)+=-,                   ①
y₀=-=-.                       ②
由①②得p=2.
(2)設(shè)N(x,y),A,B,
x₁≠x₂,由N為線段AB中點(diǎn)知
x=,                                       ③
y=.                                       ④
切線MA,MB的方程為
y=(x-x₁)+  ,                                 ⑤
y=(x-x₂)+  .                                 ⑥
由⑤⑥得MA,MB的交點(diǎn)M(x₀,y₀)的坐標(biāo)為
x₀=,y₀=.
因?yàn)辄c(diǎn)M(x₀,y₀)在C₂上,
即=-4y₀,
所以x₁x₂=-.                                ⑦
由③④⑦得
x²=y,x≠0.
當(dāng)x₁=x₂時(shí),A,B重合于原點(diǎn)O,AB中點(diǎn)N為O,坐標(biāo)滿足x²=y.
因此AB中點(diǎn)N的軌跡方程為x²=y.