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2013年陕西省高考数学理科卷第20题是:已知动圆过定点A(4,0),且在y轴上截得的弦MN的长为8.(Ⅰ)求动圆圆心的轨迹C的方程;(Ⅱ)已知点B(-1,0),设不垂直于x轴的直线l与轨迹C交于不同的两点P,Q.若x轴是∠PBQ的角平分线,证明直线l过定点.解析(Ⅰ)设动圆圆心C的坐标为(x,y),则(4-x)~2+(0-y)~2=4~2+x~2.整理得,y~2=8x.故所求动圆圆心的轨迹C的方程为y~2=8x.
In 2013, the 20th item of mathematics science volume of Shaanxi province college entrance examination is: It is known that the moving point A (4,0) is fixed and the length of the string MN taken on the y axis is 8. (Ⅰ) (2) Known point B (-1, 0), set the line not perpendicular to the x-axis and the intersection of the trajectory C and the two different points P, Q. If the x-axis is the angle bisector of ∠PBQ , Prove that the line l is over a fixed point. Analysis (Ⅰ) Set the coordinates of the circle center C as (x, y), then (4-x) ~ 2 + (0-y) ~ 2 = 4 ~ 2 + x ~ 2. Finishing, y ~ 2 = 8x. So seeking to move the center of the circle trajectory C equation is y ~ 2 = 8x.