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今年高校统考数学试卷第九题: 给定双曲线x~2-y~2/2=1, (1)过点A(2,1)的直线与所给双曲线交于两点P_1及P_2,求线段P_1P_2的中点P的轨迹方程。 (2)过点B(1,1)能否作直线m,使m与所给双曲线交于两点Q_1及Q_2,且点B是线段Q_1Q_2的中点?这样的直线如果存在,求出它的方程;如果不存在,说明理由。解这一类问题,一般是联立曲线方程得方程组,化为一元二次方程,利用韦达定理,而不必求出交点坐标。解:(1)设各点坐标为P_1(x_1,y_1)、P_2(x_2,y_2)、P(x,y),又设过点A(2,1)的直线1的方程为y-1=k(x-2),即y=kx+(1-2k),与
This year's college math test paper ninth question: given hyperbolic x ~ 2-y ~ 2/2 = 1, (1) point A (2,1) and the given hyperbolas at two points P_1 and P_2 Find the trajectory equation of the midpoint P of the segment P_1P_2. (2) Can B (1,1) be a straight line m such that m and the given hyperbolas intersect at two points Q_1 and Q_2, and point B is the midpoint of segment Q_1Q_2? If such a straight line exists, find Its equation; if not, explain the reason. Solve this type of problem, the general equation is a set of simultaneous equations, into a quadratic equation, the use of Veda's theorem, without having to find the intersection coordinates. Solution: (1) Let the equation of the straight line 1 with coordinates of P_1 (x_1, y_1), P_2 (x_2, y_2), P (x, y) = k (x-2), that is, y = kx + (1-2k), and