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探求曲线的轨迹方程,即求曲线上动点坐标所满足的代数条件是解析几何的最基本问题,它在历年高考中频繁出现.此类问题一般是通过建立坐标系,设动点坐标,依据题设条件,列出等式,代入化简整理即得曲线的轨迹方程.现结合近年的高考试题,介绍几种常用方法.一、直接法若动点运动过程中量的关系简明,那么直接将此量的关系坐标化,列出等式,化简即得动点的轨迹方程.例1已知直角坐标平面上一点 Q(2,0)和圆 C:x~2+y~2=1,动点 M 到圆 C 的切线长等于圆C 的半径与|MQ|的和,求动点 M的轨迹方程,说明它表示什么曲线,并画出草图(1994年全国高考题).
The search for the trajectory equation of the curve, that is, the algebraic condition satisfied by the coordinate of the moving point on the curve is the most basic problem of analytical geometry. It occurs frequently in the college entrance examinations over the years. Such problems are generally established by establishing a coordinate system and setting the coordinates of the moving point. Set the conditions, list the equations, and substitute the trajectory equations to simplify the curve. Now combine several high-level examination questions, introduce several common methods. First, the direct method if the relationship between the amount of movement in the moving point is concise, then direct Coordinate this relationship, list the equations, and simplify the trajectory equation of the resulting point. Example 1 Known point Q(2,0) and circle C:x~2+y~2= in the Cartesian coordinate plane 1, the tangent length of the moving point M to the circle C is equal to the radius of the circle C and the sum of |MQ|, the trajectory equation of the seeking point M, indicating what curve it represents, and drawing a sketch (the national college entrance examination question in 1994).