高考数学问题:已知抛物线y^2=4x的准线与x轴交于M点

1个回答

  • 抛物线C:y^2=4x的准线:x=-1,M(-1,0)

    AB:y=kx+k,x=(y-k)/k,设AB的中点E

    y^2=4x=4*(y-k)/k

    ky^2-4y+4k=0

    yA+yB=4/k,yA*yB=4

    (yA-yB)^2=(yA+yB)^2-4yA*yB=(4/k)^2-4*4=(16-16k^2)/k^2

    yA=yB,则AB与C相切

    yA≠yB, (yA-yB)^2>0,16-16k^2>0,|k|3

    (2)△ABD是正△,DE=AB*cos30°

    (xA-xB)^2=(yA-yB)^2/k^2

    AB^2=(1+1/k^2)*(yA-yB)^2=[(1+k^2)/k^2]*(16-16k^2)/k^2=16(1-k^4)/k^4

    DE^2=(xE-xD)^2+(yE)^2=[(2-k^2)/k^2-(2+k^2)/k^2]^2+(2/k)^2=4(1+k^2)/k^2

    DE^2=AB^2*(cos30°)^2

    4(1+k^2)/k^2=[16(1-k^4)/k^4]*3/4

    (4k^2-3)*(k^2+1)=0

    k^2=3/4

    x0=(2+k^2)/k^2=11/3

    x0=11/3,三角形ABD是正三角形.