Property for Conjugate diameters in an ellipse

Question

About six months ago I came up with a nice property for Conjugate diameters in an ellipse but couldn't prove it

If anyone can prove this, please do so

If the feature is already discovered, please point to a reference mentioning it

Answer 1

I'll prove this property from scratch.

Let $ t_0 \in [0, \dfrac{\pi}{2} ] $, and let $c = \cos t_0 $ and $s = \sin t_0$

Let $a$ be the semi-major axis length, and $b$ the semi-minor axis length.

Then

$A = (a c, b s) $

$B = (-a s, b c) $

$F = (a e , 0 )$

$F' = (- a e, 0 )$

where $e = \sqrt{1 - \left(\dfrac{b}{a}\right)^2 }$

Then,

$AF = (a (e - c) , - b s)$

$AF' = (a (-e - c) , - b s )$

$BF = ( a (e + s ) , - b c ) $

$BF' = ( a (-e + s) , - b c ) $

Now the squared lengths are

$|AF|^2 = a^2(e - c)^2 + b^2 s^2 $

$ |AF'|^2 = a^2 (e + c)^2 + b^2 s^2 $

$ |BF|^2 = a^2 (e + s)^2 + b^2 c^2 $

$ | BF' |^2 = a^2 (e - s)^2 + b^2 c^2 $

Now the product of the first two is

$ |AF|^2 |AF'|^2 = ( a^2 (e^2 + c^2) + b^2 s^2 )^2 - ( 2 a^2 e c )^2 \\ =a^4 (e^4 + 2 e^2 c^2 + c^4) + (b^2 s^2)^2 + 2 a^2 b^2 s^2 (e^2 + c^2) - 4 a^4 e^2 c^2 \\ = a^4 (e^4 - 2 e^2 c^2 + c^4) + (b^2 s^2)^2 + 2 a^2 b^2 s^2 (e^2 + c^2) \\ = ( a^2 (e^2 - c^2) )^2 + (b^2 s^2)^2 + 2 a^2 b^2 s^2 (e^2 + c^2) $

Substitute $a^2 e^2 = a^2 - b^2$ , then

$|AF|^2 |AF'|^2 = ( a^2 - b^2 - a^2 c^2 )^2 + (b^2 s^2)^2 + 2 b^2 s^2 (a^2 - b^2 + a^2 c^2) \\ = (a^2 s^2 - b^2)^2 + b^4 s^4 + 2 b^2 s^2 (a^2 - b^2 + a^2 c^2)\\ = a^4 s^4 + b^4 - 2 a^2 b^2 s^2 + b^4 s^4 + 2 a^2 b^2 s^2 - 2 b^4 s^2 + 2 a^2 b^2 s^2 c^2\\ = a^4 s^4 + b^4 + b^4 s^4 - 2 b^4 s^2 + 2 a^2 b^2 s^2 c^2 \\ = (a^2 s^2 + b^2 c^2)^2 + b^4 - b^4 c^4 + b^4 s^4 - 2 b^4 s^2 \\ = (a^2 s^2 + b^2 c^2)^2 + b^4 + b^4 ( s^4 - c^4 - 2 s^2 ) \\ = (a^2 s^2 + b^2 c^2)^2 + b^4 + b^4 ( s^4 - (1 - s^2)^2 - 2 s^2 ) \\ = (a^2 s^2 + b^2 c^2)^2 $

Switching $s$ and $c$ gives us

$ |BF|^2 |BF'|^2 = (a^2 c^2 + b^2 s^2)^2 $

The product of the two expressions is

$ (|AF||AF'||BF||BF'|)^2 = (a^2 s^2 + b^2 c^2 )^2 (a^2 c^2 + b^2 s^2)^2 $

So that

$ |AF||AF'||BF||BF'| = (a^2 s^2 + b^2 c^2) (a^2 c^2 + b^2 s^2) $

But,

$|OA|^2 = |A|^2 = a^2 c^2 + b^2 s^2 $

and

$|OB|^2 = |B|^2 = a^2 s^2 + b^2 c^2 $

Therefore, we have shown that

$ |AF||AF'||BF||BF'| = |OA|^2 |OB|^2 $

Answer 2

In fact, your property comes from this "more elementary" property applied twice, firstly for $M=A$, then for $M=B$ :

Property 1 : "Let $M$ be a point on an ellipse with foci $F,F'$ and center $O$. The product $MF \times MF'$ is equal to the square $OM'^2$ of the length of the semi-diameter conjugate to the diameter $OM$ passing through point $M$".

(this property is mentionned for example at the bottom of page 7 in this comprehensive list of results).

Here is a proof of property 1 above using an interesting characterization of conjugate diameters :

Property 2 : The eccentric angles of the ends of a pair of conjugate diameters differ by $\tfrac{\pi}{2}$. Said in a different way :

$$\vec{OP}=(a \cos \theta, b \sin \theta) \ \ \text{has}$$

$$\vec{OP'}=(a \cos(\theta+\tfrac{\pi}{2}),b \sin(\theta+\tfrac{\pi}{2})$=(-a \sin \theta,b \cos \theta)$$

as one of its 2 conjugate vectors.

Besides, the coordinates of $F,F'$ are

$$F,F'=(\pm f, 0) \ \text{with} \ f^2=a^2-b^2$$

We have to prove that :

$$MF^2 \times MF'^2 = OM'^4$$

$$\left((a\cos \theta-f)^2+b^2 (\sin \theta)^2)\right)\left((a\cos \theta+f)^2+b^2 (\sin \theta)^2\right)=(a^2 (\sin \theta)^2+b^2 (\cos \theta)^2)^2$$

for all $\theta$, which amounts to prove that :

$$(a^2 \cos^2 \theta + b^2 \sin^2 \theta + f^2)^2-4a^2f^2 \cos^2 \theta=(a^2 \sin^2 \theta + b^2 \cos^2 \theta)^2$$

which is routine verification (by replacing for example $(\sin \theta)^2$ by $1-(\cos \theta)^2$ and $f^2$ by $a^2-b^2$.