已知平面两定点A(1,3),B(4,2).试在方程为x^2/9+y^2/4=1（x,y均≥0）的椭圆周上求一点C,使△ABC的面积最用拉格朗日.

Question

已知平面两定点A(1,3),B(4,2).试在方程为x^2/9+y^2/4=1（x,y均≥0）的椭圆周上求一点C,使△ABC的面积最
用拉格朗日.

Answer 1

AB=√[(4-1)^2+(2-3)^2]=√10
k(AB)=(2-3)/(4-1)=-1/3
AB直线方程为：y-3=-1/3*(x-1)
整理得 x+3y-10=0
设椭圆参数方程为x=3cost,y=2sint,0≤t≤π/2
设椭圆上点C=C(3cost,2sint)
则点C到AB的距离为d=|3cost+6sint-10|/√10
S△ABC=1/2*AB*d
=1/2*√10*|3cost+6sint-10|/√10
=1/2*|3cost+6sint-10|
∵0≤t≤π/2,∴0≤sin≤1,0≤cost≤1,∴3cos+6sint-10<0
∴S△ABC=1/2*[10-3(2sint+cost)]
=1/2*[10-3√5sin(t+u)]        (tanu=1/2,u≈0.46=0.15π)
≥1/2*[10-3√5]
∴当sin(t+u)=1时,S△ABC取得最小值(10-3√5)/2
此时,t+u=π/2,∴t=π/2-u=0.35π
sin0.35π≈0.89,cos0.35π≈0.45
∴点C坐标为C(3cos0.35π,2sin0.35π)=C(1.35,1.78)