Question

已知椭圆中心为O，长轴、短轴的长分别为2a,2b(a>b>0),A,B分别为椭圆上的两点，且OA垂直OB· (1)1/|OA|2...

已知椭圆中心为O，长轴、短轴的长分别为2a,2b(a>b>0),A,B分别为椭圆上的两点，且OA垂直OB·
(1)1/|OA|2+1/|OB|2定值。
(2)求△AOB面积的最大值和最小值。
急求，谢谢了！

Answer 1

设OA的所在直线方程为y=kx,则OB所在直线方程为y=-x/k;
它们与椭圆的交点A、B坐标（xa,ya)、(xb,yb)满足
xa^2=1/[1/a^2+k^2/b^2]
ya^2=k^2/[1/a^2+k^2/b^2]
xb^2=1/[1/a^2+1/(k^2b^2)]
yb^2=1/[k^2/a^2+1/b^2]
OA^2=xa^2+ya^2=(1+k^2)/[1/a^2+k^2/b^2]
OB^2=xb^2+yb^2=(1+1/k^2)/[1/a^2+1/(k^2b^2)]
1/OA^2+1/OB^2=[1/a^2+k^2/b^2]/(1+k^2)+[1/a^2+1/(k^2b^2)]*k^2/(1+k^2)
=1/a^2+1/b^2为定值。

=======================================================
以中心为极点，x轴为极轴建立极坐标系
方程为ρ^2(cosθ)^2/a^2+ρ^2(sinθ)^2/b^2=1
1/ρ^2=(cosθ)^2/a^2+(sinθ)^2/b^2
设A(ρ1,θ)，由OA⊥OB得B(ρ2,θ+π/2)
1/OA^2+1/OB^2＝1/ρ1^2+1/ρ2^2
=(cosθ)^2/a^2+(sinθ)^2/b^2+(cos(θ+π/2))^2/a^2+(sin(θ+π/2))^2/b^2
=(cosθ)^2/a^2+(sinθ)^2/b^2+(sinθ)^2/a^2+(cosθ)^2/b^2
=1/a^2+1/b^2

(2)S=1/2|OA|*|OB|<=1/2(OA^2+OB^2)/2

Answer 2

设OA的所在直线方程为y=kx,则OB所在直线方程为y=-x/k;
它们与椭圆的交点A、B坐标（xa,ya)、(xb,yb)满足
xa^2=1/[1/a^2+k^2/b^2]
ya^2=k^2/[1/a^2+k^2/b^2]
xb^2=1/[1/a^2+1/(k^2b^2)]
yb^2=1/[k^2/a^2+1/b^2]
OA^2=xa^2+ya^2=(1+k^2)/[1/a^2+k^2/b^2]
OB^2=xb^2+yb^2=(1+1/k^2)/[1/a^2+1/(k^2b^2)]
1/OA^2+1/OB^2=[1/a^2+k^2/b^2]/(1+k^2)+[1/a^2+1/(k^2b^2)]*k^2/(1+k^2)
=1/a^2+1/b^2为定值。

=======================================================
以中心为极点，x轴为极轴建立极坐标系
方程为ρ^2(cosθ)^2/a^2+ρ^2(sinθ)^2/b^2=1
1/ρ^2=(cosθ)^2/a^2+(sinθ)^2/b^2
设A(ρ1,θ)，由OA⊥OB得B(ρ2,θ+π/2)
1/OA^2+1/OB^2＝1/ρ1^2+1/ρ2^2
=(cosθ)^2/a^2+(sinθ)^2/b^2+(cos(θ+π/2))^2/a^2+(sin(θ+π/2))^2/b^2
=(cosθ)^2/a^2+(sinθ)^2/b^2+(sinθ)^2/a^2+(cosθ)^2/b^2
=1/a^2+1/b^2

(2)S=1/2|OA|*|OB|<=1/2(OA^2+OB^2)/2

Answer 3

A的所在直线方程为y=kx,则OB所在直线方程为y=-x/k;
它们与椭圆的交点A、B坐标（xa,ya)、(xb,yb)满足
xa^2=1/[1/a^2+k^2/b^2]
ya^2=k^2/[1/a^2+k^2/b^2]
xb^2=1/[1/a^2+1/(k^2b^2)]
yb^2=1/[k^2/a^2+1/b^2]
OA^2=xa^2+ya^2=(1+k^2)/[1/a^2+k^2/b^2]
OB^2=xb^2+yb^2=(1+1/k^2)/[1/a^2+1/(k^2b^2)]
1/OA^2+1/OB^2=[1/a^2+k^2/b^2]/(1+k^2)+[1/a^2+1/(k^2b^2)]*k^2/(1+k^2)
=1/a^2+1/b^2为定值。

=======================================================
以中心为极点，x轴为极轴建立极坐标系
方程为ρ^2(cosθ)^2/a^2+ρ^2(sinθ)^2/b^2=1
1/ρ^2=(cosθ)^2/a^2+(sinθ)^2/b^2
设A(ρ1,θ)，由OA⊥OB得B(ρ2,θ+π/2)
1/OA^2+1/OB^2＝1/ρ1^2+1/ρ2^2
=(cosθ)^2/a^2+(sinθ)^2/b^2+(cos(θ+π/2))^2/a^2+(sin(θ+π/2))^2/b^2
=(cosθ)^2/a^2+(sinθ)^2/b^2+(sinθ)^2/a^2+(cosθ)^2/b^2
=1/a^2+1/b^2

(2)S=1/2|OA|*|OB|<=1/2(OA^2+OB^2)/2

Answer 4

http://zhidao.baidu.com/question/95345611.html

Answer 5

设OA的所在直线方程为y=kx,则OB所在直线方程为y=-x/k;
它们与椭圆的交点A、B坐标（xa,ya)、(xb,yb)满足
xa^2=1/[1/a^2 k^2/b^2]
ya^2=k^2/[1/a^2 k^2/b^2]
xb^2=1/[1/a^2 1/(k^2b^2)]
yb^2=1/[k^2/a^2 1/b^2]
OA^2=xa^2 ya^2=(1 k^2)/[1/a^2 k^2/b^2]
OB^2=xb^2 yb^2=(1 1/k^2)/[1/a^2 1/(k^2b^2)]
1/OA^2 1/OB^2=[1/a^2 k^2/b^2]/(1 k^2) [1/a^2 1/(k^2b^2)]*k^2/(1 k^2)
=1/a^2 1/b^2为定值。

=======================================================
以中心为极点，x轴为极轴建立极坐标系
方程为ρ^2(cosθ)^2/a^2 ρ^2(sinθ)^2/b^2=1
1/ρ^2=(cosθ)^2/a^2 (sinθ)^2/b^2
设A(ρ1,θ)，由OA⊥OB得B(ρ2,θ π/2)
1/OA^2 1/OB^2＝1/ρ1^2 1/ρ2^2
=(cosθ)^2/a^2 (sinθ)^2/b^2 (cos(θ π/2))^2/a^2 (sin(θ π/2))^2/b^2
=(cosθ)^2/a^2 (sinθ)^2/b^2 (sinθ)^2/a^2 (cosθ)^2/b^2
=1/a^2 1/b^2

(2)S=1/2|OA|*|OB|<=1/2(OA^2 OB^2)/2