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例题如图1,已知椭圆x~2/2+y~2=1的右准线l与x轴相交于E,过椭圆的右焦点F的直线与椭圆相交于A、B两点,点C在右准线l上,且BC∥x轴。求证:直线AC经过线段EF的中点。证明:椭圆的右准线为x=2,E(2,0),F(1,0)。所以EF的巾点N的坐标为(3/2,0)。若AB垂直于x轴,显然AC过EF的中点。若AB不垂直于x轴,设AB:y=k(x-1)(k≠0),将其代人椭圆方程得(1+
Example of the problem shown in Figure 1, known elliptical x ~ 2/2 + y ~ 2 = 1 Right alignment l intersects the x-axis at E, oval right focus F intersects the ellipse and ellipse at two points A, B C is on the right line l, and BC // x axis. Verify: The line AC passes through the midpoint of segment EF. Proof: The right line of the ellipse is x = 2, E (2,0), F (1,0). Therefore, the coordinates of the napkin N of EF are (3 / 2,0). If AB is perpendicular to the x-axis, it is clear that AC crosses the midpoint of EF. If AB is not perpendicular to the x-axis, let AB: y = k (x-1) (k ≠ 0)