Let p (x0, y0) = = > Equation x0 2/4+y0 2 =1,L: x0 x/4+y0y =1= > | OA | = 4/| x0 ||| =/.

Because of symmetry, x0>0, y0>0 = = =>s triangle OAB = | OA |||| OB |/2 = 2/(x0y0).

x0^2/4+y0^2= 1===>； 1 & gt； = 2 * x0/2 * y0 = x0 y0 & gt； 0 = = = & gt 1/(x0y0)>= 1 = = = >； S & gt=2.