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例1一动点到定直线x=-4的距离是它到定点F(-5,0)的距离的三分之一,求动点的轨迹方程.错解:因为动点到定点的距离是它到定直线距离的三倍,即e=3>1,所以动点的轨迹是双曲线,且焦点是F(-5,0),对应准线是X=-4.∵-c=-5,-a~2/c=-4.∴a~2=20,b~2=c~2-a~2=5∴动点的轨迹方程为:x~2/20-y~2/5=1剖析:由e=3>1,知动点轨迹是双曲线正确,但中心是否在坐标原点,题目没有给出,上述解答的错误在于添加条件:中心在坐标原点.
Example 1 The distance from a fixed point to a fixed line x=-4 is one-third of its distance from the fixed point F(-5,0). The trajectory equation of the seeking point. Mistake: Because the distance from the fixed point to the fixed point is It is three times the straight line distance, ie, e=3>1, so the trajectory of the moving point is hyperbolic, and the focus is F(-5,0), and the corresponding guideline is X=-4.∵-c=- 5,-a~2/c=-4.∴a~2=20,b~2=c~2-a~2=5 The trajectory equation of the inflection point is: x~2/20-y~2/ 5=1 analysis: from e=3>1, the trajectory of the kinematic point is correct, but the center is not in the origin of the coordinate. The error of the above solution lies in adding the condition: the center is at the origin of the coordinate.