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例题抛物线y~2=2px(p>0)的顶点是O,A,B是抛物线上两个异于O的动点,且满足OA⊥OB,证明:直线AB过一个定点.分析要证明动直线AB过一个定点,首先要求出直线AB的方程,而要求方程需先找到两个点的坐标,于是我们可以利用两条动直线来求交点.易知直线OA,OB的斜率都存在,设直线OA的方程为y=kx,将其代入y2=2px,可解得点A的坐标.同理,设直线OB的方程为y=-1/kx,同样可解得点B的坐标,由此可求出直线的方程.
The vertices of the example parabola y ~ 2 = 2px (p> 0) are O, A and B are two moving points different from 0 on the parabola and satisfy OA ⊥ OB, proving that the straight line AB crosses a fixed point. Straight line AB over a fixed point, first of all require a straight line AB equation, and require equations need to find the coordinates of the two points, so we can use two moving straight lines to find the intersection .I know the straight line OA, OB slope exists, The equation of a straight line OA is y = kx, substituting it into y2 = 2px can solve the coordinate of point A. Similarly, suppose the equation of a straight line OB is y = -1 / kx, and the coordinates of point B can be solved as well Find a straight line equation.