1. 設(shè)P點(diǎn)坐標(biāo)為(x,y)
MN=(-3, 0)
MP=(x-4, y)
NP=(x-1, y)
由已知(-3)*(x-4)+0*y=6*√[(x-1)²+y²]
平方 (x-4)²=4(x-1)²+4y²
化簡得3x²+4y²=12
即x²/4+y²/3=1
為一個(gè)橢圓
2. 設(shè)過點(diǎn)N的直線L為y=k(x-1)
代入橢圓方程 3x²+4k²(x-1)²=12
(3+4k²)x²-8k²x+4k²-12=0
設(shè)A(x1,y1)B(x2,y2)
則x1,x2 是上式的兩個(gè)根
由韋達(dá)定理x1+x2=8k²/(4k²+3) x1*x2=(4k²-12)/(4k²+3)
所以(1-x1)(x2-1)=(x1+x2)-x1*x2-1
=8k²/(4k²+3)-(4k²-12)/(4k²+3)-1
=(8k²-4k²+12-4k²-3)/(4k²+3)
=9/(4k²+3)
將A,B代入直線L:y1=k(x1-1) , y2=k(x2-1)
則y1*y2=k²(x1-1)(x2-1)=-k²(1-x1)(x2-1)
由已知5*向量NA*向量BN
=5*[(1-x1)(x2-1)-y1*y2]
=5[(1-x1)(x2-1)+k²(1-x1)(x2-1)]
=5(1+k²)(1-x1)(x2-1)
=5(1+k²)*9/(4k²+3)=12
即15(1+k²)=4(4k²+3)
解得k=±√3
所以直線L的方程為y=±√3(x-1)