2)y=k(x-1)与x^2/4+y^3=1联立得(4k^2+3)x^2-8k^2x+4k^2-12=0
韦达定理得x1+x2=8k^2/(4k^2+3) ∴PQ中点为(4k^2/(4k^2+3),-3k/(4k^2+3))
∴线段PQ的垂直平分线为y=(-1/k)(x-4k^2/(4k^2+3))-3k/(4k^2+3)
x=0时,y=k/(4k^2+3)=1/(4k+3/k)属于(-根号3/12,根号3/12)
3)设M(x,0) 向量MP*MQ=(x1-x,y1)*(x2-x,y2)=x1x2-x(x1+x2)+x^2+y1y2=(1+k^2)(4k^2-12)/(4k^2+3)-8(x+k^2)k^2/(4k^2+3)+x^2+k^2=[(4x^2-8x-5)k^2+3x^2-12]/(4k^2+3)为定值
∴[4x^2-8x-5]/4=(3x^2-12)/3解得x=11/8 即M(11/8,0)