求椭圆2x^2+y^2=6在点(1,2)处的切线方程.解：对2x^2+y^2=6求导得4x+2y*y'=0,y'=-2x/y,在点(1,2)处y'=-1,所以所求的切线方程是y-2=-(x-1),即x+y-3=0.
