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

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.