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过点M(11)的直线过点M(1,1)的直线,交椭圆x^2/3+y

2011-01-13 18:30:39a***
过点M(1,1)的直线,交椭圆x^2/3+y^2/2=1于A,B,求AB的中点轨迹方程。过点M(11)的直线过点M(1,1)的直线,交椭圆x^2/3+y^2/2=1于A,B,求AB的中点轨迹方程。:设A(x1,y1),B(x2,y2). AB的中点?

最佳回答

  • 设A(x1,y1),B(x2,y2). AB的中点M(x,y). ∵ 2(x1)^2+3(y1)^2=6......①,2(x2)^2+3(y2)^2=6......②, ①-②得, 2(x1+x2)(x1-x2)+3(y1+y2)(y1-y2)=0,而x1+x2=2x,y1+y2=2y,过点M(1,1)的直线AB的斜率k=(y1-y2)/(x1-x2), ∴ 2x+3ky=0,k=-2x/3y,代入y=k(x-1)+1得轨迹方程:4[x-(1/2)]^2+6y^2=1
    2011-01-13 20:44:45
  • 设过点M(1,1)的直线l:y=k(x-1)+1,① 代入x^2/3+y^2/2=1,得 (2+3k^2)x^2+6k(1-k)x+3k^2-6k-3=0, 设A(x1,y1),b(x2,y2),则 x1+x2=6k(k-1)/(2+3k^2), ∴AB的中点坐标满足 x=(x1+x2)/2=3k(k-1)/(2+3k^2), 由①,k=(y-1)/(x-1), 代入上式,化简得 x=3(y-1)(y-x)/[2(x-1)^2+3(y-1)^2], x(2x^2+3y^2-4x-6y+5)=3(y^2-xy-y+x), ∴2x^3+3xy^2-4x^2-3xy-3y^2+2x+3y=0,为所求。
    2011-01-13 20:10:24
  • 很赞哦! (34)