解:1.设动圆的圆心为(x,y)圆心到定点的距离与到切线的距离相等,即√[(x-p/2)^2+y^2]=|x+p/2|两边平方并整理得:y^2=2px,即动圆圆心的轨迹C的方程为y^2=2px (p>0)；2.直线OA的方程:y=(tanα)x直线OB的方程:y=(tanβ)x分别代入y^2=2px中得A的坐标:(2p/(tanα)^2,2p/(tanα))B的坐标:(2p/(tanβ)^2,2p/(tanβ))直线AB的斜率:k=[2p/tanα-2p/tanβ]/[2p/(tanα)^2-2p/(tanβ)^2]整理得:k=(tanαtanβ)/(tanα+tanβ)直线AB的方程:y-2p/tanα=[(tanαtanβ)/(tanα+tanβ)](x-2p/(tanα)^2)整理得:(tanαtanβ)x-(tan+tanβ)y+2p=0 (*)tanθ=tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)所以tanα+tanβ=tanθ-tanθtanαtanβ 代入(*)中得(tanαtanβ)x-tanθy+tanθtanαtanβy+2p=0(tanαtanβ)(x+tanθy)-tanθy+2p=0令x+tanθy=0,-tanθy+2p=0得到x=-2p,y=2p/tanθ即直线AB恒过定点(-2p,2p/tanθ)