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题目点A为y轴正半轴上一点,A、B两点关于x轴对称,过点A任作直线交抛物线y=2/3x~2于P、Q两点.(1)求证:∠ABP=∠ABQ.(2)若点A的坐标为(0,1),且∠PBQ=60°,试求所有满足条件的直线PQ的函数解析式(如图).文[1]利用轴对称知识及函数与方程思想进行解答,应该肯定解法很全新,笔者本着一切从学生所掌握的基本知识出发来解答,从三角形角平分线定理入手,解答比较通俗简单,供同学们参考.(1)证明设点A坐标为(0,a),P、Q坐标分别为(x_1,y_1)、(x_2,y_2),令直线PQ方程:y=kx+a,再联立y=2/3x~2解得2/3x~2-kx-a=0,则x_1x_2=-3/2a(即a=-2/3x_1x_2),y_1=2/3x_1~2、
The subject point A is a point on the positive axis of the y-axis, and the two points A and B are symmetrical about the x-axis. If the point A is crossed, let parabola be y = 2 / 3x ~ 2 at P and Q. (1) ABP = ∠ABQ. (2) If the coordinate of point A is (0,1), and ∠PBQ = 60 °, try to find the function analytic formula of all the straight line PQ that satisfies the condition Symmetric knowledge and function and equation thinking to answer, we should affirm the solution is very new, the author of everything from the students to master the basic knowledge to answer, starting from the triangle angle bisector theorem, the answer is more simple, for the students to reference. 1) Prove that the coordinate of point A is (0, a) and the coordinates of P and Q are (x_1, y_1) and (x_2, y_2) respectively. Let the linear PQ equation: y = kx + 3x ~ 2 for 2 / 3x ~ 2-kx-a = 0, x_1x_2 = -3 / 2a (that is a = -2/3 x_1x_2), y_1 = 2/3x_1 ~ 2,