Question

一道高中数学题，用参数方程解，求详细。。

Answer 1

由已知设椭圆方程x^2/a^2+y^2/b^2=1
(1)设|OA|=r,|OB|=R,不妨设射线OA、OB与x轴正方向所成角分别是θ、θ+π/2
则A(rcosθ,rsinθ)、B(Rcos(θ+π/2),Rsin(θ+π/2))
A在椭圆上:(rcosθ)^2/a^2+(rsinθ)^2/b^2=1
得1/r^2=(cosθ)^2/a^2+(sinθ)^2/b^2 (1)
B在椭圆上:(Rcos(θ+π/2))^2/a^2+(Rsin(θ+π/2))^2/b^2=1
(-Rsinθ)^2/a^2+(Rcosθ)^2/b^2=1
得1/R^2=(sinθ)^2/a^2+(cosθ)^2/b^2 (2)
(1)+(2):(1/r^2)+(1/R^2)=((cosθ)^2/a^2+(sinθ)^2/b^2)+((sinθ)^2/a^2+(cosθ)^2/b^2)
(1/r^2)+(1/R^2)=(1/a^2+1/b^2)((sinθ)^2+(cosθ)^2)=
(1/r^2)+(1/R^2)=1/a^2+1/b^2
即1/|OA|^2+1/|OB|^2=1/a^2+1/b^2
所以1/|OA|^2+1/|OB|^2为定值.
(2)设△OAB的面积为S,则S=(Rr)/2, 由(1)
1/r^2=(1/a^2)+(c^2/(ab)^2)(sinθ)^2 (3)
1/R^2=(1/a^2)+(c^2/(ab)^2)(cosθ)^2 (4)
(3)×(4)：1/(Rr)^2=1/a^4+(c^2/(a^4b^2)+(c^4/(ab)^4)(sinθcosθ)^2
而0≤(sinθcosθ)^2≤1/4
有1/(Rr)^2≥1/a^4+(c^2/(a^4b^2)=1/(ab)^2 ((sinθcosθ)^2=0时取"=")
且1/(Rr)^2≤1/a^4+(c^2/(a^4b^2)+(c^4/(ab)^4)(1/4)
=(1/4)(1/a^2+1/b^2)^2 ((sinθcosθ)^2=1/4时取"=")
得1/(ab)≤1/(Rr)≤(1/2)(1/a^2+1/b^2)
(ab)/2≥S≥(ab)^2/(a^2+b^2) 且两边等号都能取到.
所以△OAB的面积S的最大值是(ab)/2,最小值是(ab)^2/(a^2+b^2) .

较难的问题,希望能帮到你！

Answer 2