Question

如圖,點(diǎn)A、B分別是橢圓=1長(zhǎng)軸的左、右端點(diǎn),點(diǎn)F是橢圓的右焦點(diǎn),點(diǎn)P在橢圓上,且位于x軸上方,PA⊥PF.

(1)求點(diǎn)P的坐標(biāo);

(2)設(shè)M是橢圓長(zhǎng)軸AB上的一點(diǎn),M到直線AP的距離等于|MB|,求橢圓上的點(diǎn)到點(diǎn)M的距離d的最小值.

Answer 1

1、P(,).

2、d_最小=.

解析:

(1)由條件A(-6,0),F(4,0).

設(shè)P(x,y).

∴=(x+6,y),=(x-4,y).

∵⊥,

∴0=(x+6)(x-4)+y²,即x²+2x-24+y²=0.                                           ①

結(jié)合+=1,                                                              ②

解得x=或-6,代入①解得y=±或0.

∴P(,±).

又點(diǎn)P在z軸上方,∴所求P(,).

(2)由兩點(diǎn)式得直線AP:x-y+6=0.

(不妨取x軸上方的P點(diǎn))

設(shè)M(m,0),|MB|=6-m.

由題意6-m=,解得m=2或18(舍去).

∴M(2,0).

設(shè)橢圓上任一點(diǎn)P(x,y),

d²=MP²=(x-2)²+y²=(x-2)²+20-x²

=x²-4x+24=(x-)²+15.

∵-6≤x≤6,取x=得

d_最小²=15.此時(shí)d_最小=.