An affine property of conic sections Lately I read a book on conic sections and found followinbg interesting problem: Let $\Gamma$ be an ellipse on a plane and $F_1,F_2$ be the foci of  $\Gamma$. Given a point $P\in \Gamma$, let $A_i$ be the another intersection of $PF_i$ with  $\Gamma$, $i=1,2$. Then the author claims that the value 
$$\left|\frac{PF_1}{A_1F_1}\right|+\left|\frac{PF_2}{A_2 F_2}\right|$$
is a constant that only relies on the curve $\Gamma$, and is in fact equals to $\dfrac{2+2e^2}{1-e^2}$, where $e$ stands for the eccentricity of $\Gamma$. The author mentioned that the proof relies on a key observation that $1/PF_i+1/A_iF_i$ is a constant, but didn't offer a detailed proof for it. Now here is my question:


*

*Me myself found a computational proof, which looks very ulgy. I feel like that there should exist a more conceptual and elegant one, say based on the  harmonic theory of a conic section, but I've no idea what it exactly is. 

*Another evidence of the existance of a conceptual proof is that the property "$1/AF+1/PF$ is a constant" seems holds for an any conic section and any chord $AP$ passing through a focus $F$. But how to prove the general fact?
Thanks in advance!
 A: It is a well-known property that the semi-latus rectum of any conic section is the harmonic mean between the segments of any focal chord (see for instance:  E. H. Askwith, A course of pure geometry, p. 114). In the case of an ellipse the semi-latus rectum is given by $b^2/a$, where $a$ and $b$, as usual, are the ellipse semiaxes. Hence:
$$
{1\over PF_1}+{1\over A_1F_1}={2a\over b^2},
\quad\text{that is:}\quad
{PF_1\over A_1F_1}={2a\over b^2}PF_1-1,
\quad\text{and:}\quad
{PF_2\over A_2F_2}={2a\over b^2}PF_2-1.
$$
From the last two equalities we then get
$$
{PF_1\over A_1F_1}+{PF_2\over A_2F_2}={2a\over b^2}(PF_1+PF_2)-2={4a^2\over b^2}-2.
$$
Taking into account that $e^2=1-b^2/a^2$ you can then obtain the given result.
EDIT.
Let me reproduce here Askwith's proof, which is remarkable as it works for any conic section. Let $S$ be a focus and $RXM$ the related directrix of the conic, $PQ$ a focal chord, $SL$ the semi-latus rectum (see diagram). By definition of directrix we know that
$$
{PS\over PM}={QS\over QR}={LS\over SX}=e\quad 
\text{(eccentricity of the conic section).}
$$

We have then:
$$
{1\over PS}+{1\over QS}={1\over e}\bigg({1\over PM}+{1\over QR}\bigg)=
{1\over LS}\bigg({SX\over PM}+{SX\over QR}\bigg)=
{1\over LS}\bigg({PM-SN\over PM}+{QR+KS\over QR}\bigg),
$$
that is:
$$
\tag{1}
{1\over PS}+{1\over QS}=
{1\over LS}\bigg(2-{SN\over PM}+{KS\over QR}\bigg).
$$
But, on the other hand:
$$
{SN\over PM}=e{SN\over PS}=e{KS\over QS}={KS\over QR}.
$$
It follows that the last two terms in equation $(1)$ cancel out, and it then reduces to
$$
{1\over PS}+{1\over QS}=
{2\over LS},
$$
as it was to be proved.
This proof must be slightly modified if $P$ and $Q$ lie on opposite branches of a hyperbola.
A: As @intelligenti's answer indicates, the relation follows quickly from some straightforward algebra based on the harmonic mean property. I thought I'd show some associated geometry, which helps give meaning to the resulting value.

Let focus $R$ lie on chord $\overline{PQ}$, and let $P_0$, $Q_0$, $R_0$ be the projections of these points onto the corresponding directrix. The harmonic mean property of the semilatus rectum (which I prove in this answer) is equivalent to the fact that $|RR_0|$ is the harmonic mean of $|PP_0|$ and $|QQ_0|$ (since the lengths in one set of segments are just "eccentricity-times" the lengths of the others). By the "crossed ladders" construction, $\overline{PQ_0}$ and $\overline{QP_0}$ meet at the midpoint $M$ of $\overline{RR_0}$. Projecting $M$ to $N$ on $\overline{PP_0}$ creates parallelogram $\square P_0MRN$, and thus also similar triangles $\triangle PNR$ and $\triangle RMQ$. Likewise, for elements (indicated with primes ("$\,'\,$") in the figure, and taking $P=P'$) relative to the other focus and directrix.
Note that $\overline{RR_0}\cong\overline{R'R_0'}$, and that their common length is a constant for the ellipse.

Dividing the sum in question by $2$ to avoid an unsightly $2$ in the final value, we can write:
$$\frac12\left(\frac{|PR|}{|RQ|}+\frac{|P'R'|}{|R'Q'|}\right)=\frac12\left(\frac{|PN|}{|RM|}+\frac{|P'N'|}{|R'M'|}\right)=\frac12\,\frac{|PN|+|P'N'|}{|RM|}=\frac{\frac12|NN'|}{|RM|}=\frac{|OM|}{|RM|} \tag{1}$$
This shows that the sum is indeed a constant. We can rewrite that constant thusly:
$$\frac{|OM|}{|RM|}=\frac{\frac12(|OR_0|+|OR|)}{\frac12(|OR_0|-|OR|)}=\frac{\;1+\dfrac{|OR|}{|OR_0|}\;}{\;1-\dfrac{|OR|}{|OR_0|}\;}=\frac{1+e^2}{1-e^2} \tag{2}$$
where $e$ is the ellipse's eccentricity. The final form follows from the fact that, if $a$ is the major radius, then $|OR|=ae$ and $|OR_0|=a/e$. $\square$
