這種題的癥結(jié)往往不是會(huì)不會(huì)做,而是算不算的出來(lái) - -

設(shè)A(x1,y1);B(x2,y2);M(x0,y0)
設(shè)直線AB: y-1 = k(x-2) 代入橢圓 4x^2+9y^2=36

4x^2+9[ kx-(2k-1)]^2=36→ (4+9k^2)x^2 - 18(2k-1)kx+9(2k-1)^2-36=0

所以 x0 = (x1+x2)/2 = 9(2k-1)k/(4+9k^2) 將 (y0-1)/(x0-2) =k 代入

x0 = 9(y0-1)(2y0-x0)/[4(x0-2)^2+9(y0-1)^2]

4x0(x0-2)^2+9x0(y0-1)^2 -9(y0-1)(2y0-x0) = 0
→4x0(x0-2)^2+9(y0-1)[ x0y0-x0-2y0+x0]=0
→4x0(x0-2)^2+9(y0-1)y0(x0-2)=0
→ 4x0(x0-2) + 9 (y0-1)y0=0
→ 軌跡方程為 4x^2-8x+9y^2-9=0