Question

椭圆中心O,长轴,短轴分别为2a,2b,A.B分别为椭圆的两点,OA垂直OB,求...

椭圆中心O,长轴,短轴分别为2a,2b,A.B分别为椭圆的两点,OA垂直OB,求证1/OA的模平方+1/OB的模平方为定值

Answer 1

将椭圆方程改写为：x=acosθ,y=bsinθ,其中θ为OP(x,y)与Ox轴的夹角
设A(x1,y1)对应的是θ1,B(x2,y2)对应的是θ2
根据题意,OA⊥OB,则|θ2-θ1|=π/2
不失一般性,可另θ2=θ1+π/2
则cosθ2=-sinθ1,sinθ2=cosθ1
x1
=
acosθ1,y1
=
bsinθ1；
x2
=
acosθ2
=
-asinθ1,y2
=
bsinθ2
=
bcosθ1
|OA|^2
=
x1^2
+
y1^2
=
a^2cos^2θ1
+
b^2sin^2θ1
|OB|^2
=
x2^2
+
y2^2
=
a^2sin^2θ1
+
b^2cos^2θ1
|OA|^2+|OB|^2
=
(a^2+b^2)*(cos^2θ1+sin^2θ1)
=
a^2+b^2
|OA|^2*|OB|^2
=
(a^2cos^2θ1
+
b^2sin^2θ1)*(a^2sin^2θ1
+
b^2cos^2θ1)
=
(a^4+b^4)*sin^2θ1cos^2θ1
+
a^2b^2*(cos^4θ1+sin^4θ1)
=
(a^4+b^4-2a^2b^2)*sin^2θ1cos^2θ1
+
a^2b^2*(cos^4θ1+sin^4θ1+2sin^2θ1cos^2θ1)
=
(a^2-b^2)^2*sin^2θ1cos^2θ1
+
a^2b^2*(cos^2θ1+sin^2θ1)^2
=
(a^2-b^2)^2*sin^2θ1cos^2θ1
+
a^2b^2
=
(ab)^2
+
(c*sinθ1cosθ1)^2
1/|OA|^2
+
1/|OB|^2
=
(|OA|^2
+
|OB|^2)/(|OA|^2*|OB|^2)
=
(a^2+b^2)/[(ab)^2+(c*sinθ1cosθ1)^2]
似乎不为常数嘛