已知椭圆C的长轴长是短轴长的2倍

1个回答

  • 已知椭圆C的长轴是短轴的2倍,且经过点(1,0) (1) 求椭圆C的标准方程; (2) 若过点M(0,1)的直线L

    交椭圆C(取焦点在y轴上的椭圆)于A、B,点P是线段AB的中点,当直线L绕点M旋转时,求动点P的轨迹方程

    (1)长轴在x,则长轴 = 1,短轴 = 1/2,方程:x^2 + 4y^2 = 1长轴在y,则长轴 = 2,短轴 = 1,方程:4x^2 + y^2 = 4(2)按题意,此时方程为:4x^2 + y^2 = 4L斜率存在时,设为k,则L:y = kx + 1设A(x1,y1)B(x2,y2)则4(x1)^2 + (y1)^2 = 4或4(x2)^2 + (y2)^2 = 4两个方程相减并整理可得:[(y1-y2)/(x1-x2)]·[(y1+y2)/(x1+x2)] = -4其中:[(y1-y2)/(x1-x2)] = k(AB) = k(L) = k,而P(x0,y0)是AB中点∴y1+y2 = 2y0,x1+x2=2x0∴k·(y0/x0) = -4 ,∴k = -4x0/y0 ,代入直线方程得:(y0)^2 = -4(x0)^2 + y0整理可得:4[(x0) - 0]^2 + [(y0) - (1/2)]^2 = 1/4 = (1/2)^2当L斜率不存在,即⊥x轴时,求得A、B坐标(0,2)、(0,-2),P(0,0)代入上式仍成立∴动点P的轨迹方程是以(0,1/2)为圆心,1/2为半径的圆:4x^2 + [y-(1/2)]^2 = 1/4