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2014浙江高考理数第21题:如图,设椭圆C:x2/a2+y2/b2=1(a>b>0),动直线l与椭圆C只有一个公共点P,且点P在第一象限.(1)已知直线l的斜率为k,用a,b,k表示点P的坐标.(2)若过原点O的直线l1与l垂直,证明:点P到直线l1的距离的最大值为a-b.标准答案:(1)设直线l的方程为y=kx+m(k<0),由
As shown in the figure, suppose the ellipse C: x2 / a2 + y2 / b2 = 1 (a> b> 0), the moving straight line l and the ellipse C have only one common point P, and the point P is the first (1) The slope of a straight line l is known as k, and the coordinates of point P are represented by a, b and k. (2) If the straight line l1 of the origin O is perpendicular to l, it is proved that the distance from the point P to the straight line l1 The maximum value is ab. Standard Answer: (1) Let the equation of line l be y = kx + m (k <0)