已知抛物线x^2=2py[p>0]

2个回答

  • 证明:

    抛物线x^2=2py的焦点F(0,p/2),准线方程y=-p/2

    设过焦点F的直线为y-p/2=k(x-0),即:y=kx+p/2

    点A(x1,y1),点B(x2,y2),k=(y1-y2)/(x1-x2)……(1)

    x^2=2py,y=x^2/(2p),y‘(x)=x/p

    所以过点A的切线斜率k1=y'(x1)=x1/p,过点B的切线斜率k2=y'(x2)=x2/p

    过点A的切线为:y-y1=k1(x-x1)=(x1/p)(x-x1)=x1x/p-x1^2/p=x1x/p-2y1,即:y=x1x/p-y1……(2)

    过点B的切线为:y-y2=k2(x-x2)=(x2/p)(x-x2)=x2x/p-x2^2/p=x2x/p-2y2,即:y=x2x/p-y2……(3)

    联立(1)至(3)式求得两切线的交点Q(x,y)坐标满足:

    x=p*(y1-y2)/(x1-x2)=pk… …(4)

    y=x1*pk/p-y1=kx1-y1=-p/2……(5)

    所以:点Q在抛物线x^2=2py的准线y=-p/2上.