Question

已知椭圆：x^2/3+y^2=1，过坐标原点o做两条互相垂直的射线，与椭圆分别交于A、B两点。

求三角形ABC面积的最大值，谢谢~

Answer 1

是三角形AOB面积最大值吗?
椭圆的参数方程为:
x=√3cost,
y=sint,
设A点时,x1=√3cost1,y1=sint1,
B点时,x2=√3cos(t1+π/2)=-√3sint1,
y2=sin(t1+π/2)=cost1,
|OA|=√[sint1)^2+3(cost1)^2],
|OB|=√[3(sint1)^2+(cost1)^2],
S△OAB=(1/2)√[sint1)^2+3(cost1)^2]*√[3(sint1)^2+(cost1)^2]
=(1/2)√[3(sint1)^4+3(cost1)^4+10(sint1)^2(cost1)^2]
=(1/2)√{3[(sint)^2+(cost1)^2]^2-2(sint1)^2(cost1)^2]+10(sint1)^2(cost1)^2}
=(1/2)√[(sin2t1)^2+3],
对于(sin2t1)^2,2t1=π/2,或2t1=3π/2时,绝对值最大,为1,或-1,
即t1=π/4,或t1=3π/4时,取最大值,
∴S(max)=(1/2)√(1+3)=1.
∴三角形OAB面积的最大值是1.

追问

答案不是这个

追答

设OA方程为:y=kx,
OB方程为:y=-x/k,
A(x1,y1),B(x2,y2)
椭圆方程为:x^2+3y^2=3,
x^2+3k^2x^2=3,
x1^2=3/(1+3k^2),
y1^2=3k^2/(1+3k^2),
x^2+3x^2/k^2=3,
x2^2=3k^2/(3+k^2),
y^2=3/(3+k^2),
OA^2=3/(1+3k^2)+3k^2/(1+3k^2)
=3(1+k^2(/(1+3k^2),
|OA|=√[3(1+k^2(/(1+3k^2)],
OB^2=3k^2/(3+k^2)+3/(3+k^2)
=3(1+k^2)/(3+k^2),
|OB|=√[3(1+k^2)/(3+k^2)]
S△OAB=|OA|*|OB|/2
S=(3/2)(1+k^2)/√[(1+3k^2)(3+k^2)],
∵k=tanθ,是OA的斜率,
S=(3/2)[1+(tanθ)^2]/√{[1+3(tanθ)^2][3+(tanθ)^2]}
=(3/2)(secθ)^2/√{[1+3(tanθ)^2][3+(tanθ)^2]} //*分子分母同乘(cosθ)^2
=(3/2)/√{[ (cosθ)^2+3(sinθ)^2][3(cosθ)^2+(sinθ)^2]}
=(3/2)/√{[1+2(sinθ)^2][1+2(cosθ)^2]}
=(3/2)/√[1+2+4(sinθ*cosθ)^2]
=(3/2)/√(3+(sin2θ)^2],
当分母最小时,分数最大,
当sin2θ=0时,分母最小,
∴S(max)=(3/2)/√(3+0)
=√3/2,
∴当θ=0时,S△OAB面积最大,
即只有OA=长半轴时面积最大,A、B分别是长、短.半轴的顶点。