由题意知切线的斜率不存在时x＝2为椭圆的切线,当切线的斜率存在时,设方程为y-2=k(x-2)即y=k(x-2)+2代入x^2+4y^2=4得x^2+4〔k(x-2)+2〕^2=4展开化简得（1+4k^2）x^2-16k(k-1)x+(16k^2-32k+12)=0由判别式为0得〔-16k(k...