直线斜率不存在时,l:x=2 与3x^2+4y^2=12
只有1个交点,不和题意
l有斜率时,令斜率为k l:y=k(x-2)+1
y=k(x-2)+1 与3x^2+4y^2=12联立方程组
消去y:3x²+4(kx-2k+1)²-12=0
即 (4k²+3)x²-8k(2k-1)x+4(2k-1)²-12=0
Δ =64k²(2k-1)²-4(4k²+3)[4(2k-1)²-12]>0
设A（x1,y1),B(x2,y2)
由韦达定理得：
∴x1+x2=8k(2k-1)/(4k²+3) ,
x1x2=[4(2k-1)²-12)]/(4k²+3)=8(2k²-2k-1)/(4k²+3)
向量PA=（x1-2,y1-1),向量PB=（x2-2,y2-1),
y1-1=k(x1-2) ;y2-1=k(x2-2)
向量PA●PB=(x1-2)(x2-2)+(y1-1)(y2-1)
=(x1-2)(x2-2)+k²(x1-2)(x2-2)
=(k²+1)(x1-2)(x2-2)
=(k²+1)[4-2(x1+x2)+x1x2]
=(k²+1)[4-16k(2k-1)/(4k²+3) +8(2k²-2k-1)/(4k²+3)]
=(k²+1)[（16k²+12)-16k(2k-1) +8(2k²-2k-1)]/(4k²+3)
=4(k²+1)/(4k²+3)
∵ PM^2=(2-1)²+(1-3/2)²=5/4
PA×PB=PM^2
∴4(k²+1)/(4k²+3)=5/4
∴ k²=1/4 k=±1/2
∴l:y=±1/2(x-2)+1
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