Question

高二椭圆数学题目

Answer 1

(1)x2/9+y2=1 (2)设BP：y=kx+1，BQ：y=-x/k+1 分别与x2/9+y2=1联立，得 (1/9+k2)x2+2kx=0 (1/9+1/k2)x2-2x/k=0 均已知其中一个x是B的横坐标0，故由韦达定理 x(B)+x(P)=x(P)=-2k/(1/9+k2) x(B)+x(Q)=x(Q)=(2/k)/(1/9+1/k2) S△BPQ=(1/2)|BP|·|BQ|=(1/2)√[(x(P)-0)2+(y(P)-1)2]·√[(x(Q)-0)2+(y(Q)-1)2] =(1/2)√(1+k2)·√(1+1/k2)·|x(P)|·|x(Q)|=18·√(2+k2+1/k2)/[82/9+k2+1/k2] =18|k+1/k|/[(k+1/k)2+64/9]=18/[|k+1/k|+(64/9)/|k+1/k|]≤27/8 当且仅当|k+1/k|=(64/9)/|k+1/k| 即|k+1/k|=8/3时，取"=" 所以最大面积是27/8，如果要求出k，请自行解方程3k2±8k+3=0