如果椭圆x^2/36+y^2/9=1的弦被点（4,2)平分，则弦所在的直线方程是什么？
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设弦所在的直线方程为y=kx+b ...(1),其与椭圆交于A(x1,y1),B(x2,y2)则, (x1+x2)/2 = 4, (y1+y2)/2 = 2即: x1+x2 = 8, y1+y2 = 4将(1)代入椭圆方程,化简,得:(9+36*k^2)*x^2 +72kbx +(36*b^2 -324) = 0== -72kb/(9+36*k^2) = x1+x2 = 8y1+y2 = 4 = k*(x1+x2) +2b解得: k = -1/2, b = 4因此, 设弦所在的直线方程为 y = -x/2 +4, 即: x +2y = 8