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现在使用的高级中学《平面解析几何全一册》(甲种本)教学参考书(人民教育出版社)第二章圆锥曲线的习题五第14题的解法(p.88)是有错误的,今提出我的意见。原题等腰三角形的顶点A(4,2),底边的一个端点B(3,5)。求另一个端点的轨迹方程,并说明它的轨迹是什么? 教参书中指出:所求轨迹方程是 x~2+y~2-8x-4y+10=0。 (x≠3,x≠5)。接着又说:轨迹是以A为圆心,10~(1/2)为半径的圆、但除去两点。事实上,坐标为(3,-1)和(5,5)
The solution (p.88) to problem 14 of exercise 5 in the conic section of the second chapter of the senior middle school “Frontier Analytic Geometry” (a type of textbook) (Teaching Books for People) is incorrect. This is my opinion. The vertex A(4,2) of the isosceles triangle of the original question and B(3,5) of the base edge. Find the trajectory equation of the other end point, and explain what its trajectory is? The teaching reference states: The trajectory equation is x~2+y~2-8x-4y+10=0. (x≠3, x≠5). Then said: The trajectory is a circle with a center of A and a radius of 10~(1/2), but with two points removed. In fact, the coordinates are (3,-1) and (5,5)