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How is $\frac{PF}{PD} = e = \frac{C}{A}$ ? where e is eccentricity, P stands for any point on the ellipse. $F$ stands for one of the foci. $e$ stands for eccentricity. $D$ is a point on the directrix of the ellipse. 'C' is the distance from the center to the focus of the ellipse 'A' is the distance from the center to a vertex.

This is referring to an ellipse/hyperbola/parabola and their conic sections. The problem is not the proof for how $PF/PD = e$, or how $C/A = e$, but how the two equate to each other. (The letters stem from the points/foci of an ellipse of a cone and its directrix).

What is the answer that does NOT use analytic geometry? (only trigonometry)

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    $\begingroup$ Please provide enough context for us to understand what on earth you are talking about. $\endgroup$ Apr 26, 2012 at 19:23
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    $\begingroup$ To me it's impossible to undertand your question. Could you explain what PF, PD, C, A, etc. mean? $\endgroup$ Apr 26, 2012 at 19:25
  • $\begingroup$ It sounds like a conic section problem ... ? $\endgroup$
    – Neal
    Apr 26, 2012 at 19:31
  • $\begingroup$ This is referring to an ellipse/hyperbola/parabola and their conic sections. The problem is not the proof for how PF/PD = e, or how C/A = e, but how the two equate to each other. (the letters stem from the points/foci of an ellipse of a cone and its directrix). $\endgroup$
    – Dona
    Apr 26, 2012 at 19:33

1 Answer 1

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The figure below shows a conic section (specifically an ellipse) with focus $F$ and directrix $D$. The conic is the intersection of the blue cone and the purple "cutting plane". The figure also shows the Dandelin sphere associated with $F$ and $D$. (Interestingly, the Wikipedia article mentions the focus-directrix property "can be proved" with Dandelin spheres, but doesn't give that proof!) Let's call the green "horizontal" plane containing the circle where the sphere meets the cone the Dandelin plane; call that circle the Dandelin circle.

Now, given $P$ on the conic, let $Q$ be the corresponding point on the Dandelin circle; that is, let $Q$ be the point where the segment joining $P$ to the cone apex meets the Dandelin plane. Let $R$ be the foot of the perpendicular dropped from $P$ to the Dandelin plane, and let $S$ be the foot of the perpendicular dropped from $P$ to the directrix.

conic section

(Original image credit, with description of Dandelin spheres.)

Since segments $\overline{PF}$ and $\overline{PQ}$ are both tangent to the Dandelin sphere, we must have $|\overline{PF}|=|\overline{PQ}|$. (This, by the way, is the primary magic of the Dandelin sphere.)

We can massage the focus-directrix ratio for $P$ thusly:

$$\frac{|\overline{PF}|}{|\overline{PS}|} = \frac{|\overline{PQ}|}{|\overline{PS}|}=\frac{|\overline{PR}|/(\sin\angle Q)}{|\overline{PS}|}=\frac{|\overline{PR}|}{|\overline{PS}|}\frac{1}{\sin\angle Q}=\frac{\sin \angle S}{\sin \angle Q} \tag{$\star$}$$

Clearly, $\angle S$ is constant as $P$ moves about the conic; it's the angle between the cutting plane and Dandelin plane. But $\angle Q$ is also constant: it's the ("exterior") angle that the surface (more precisely, a "generator" line) of the cone makes with the Dandelin plane. Therefore, the focus-directrix ratio is a constant.

To complete the answer to your question, all we have to do is prove that "(focal distance)-over-(major radius)" gives the same trigonometric ratio.

For now, we'll assume the conic is an ellipse (that is $\angle S$ is smaller than $\angle Q$).

Look at the figure "sideways", reducing all the elements to their intersections with the plane through the cone's axis, perpendicular to the directrix. I'll take $P$ to be the point on the conic closest to the directrix (which itself has projected into the point $S$), and $P^\prime$ the farthest point. ($Q$ and $Q^\prime$ are the corresponding points on the Dandelin plane, which has projected into a line.) Then $\overline{PP^\prime}$ is the major axis of the ellipse. The ellipse's focus, $F$, corresponds to the point where the incircle of $\triangle OPP^\prime$ meets $\overline{PP^\prime}$; the ellipse's center corresponds to $M$, the midpoint of $\overline{PP^\prime}$.

enter image description here

Now that we know where everything is, a couple more applications of the equal-tangent-segment property are all we need. With $a := |\overline{MP^\prime}|$ and $c := |\overline{MF}|$, we have

$$\frac{\sin\angle S}{\sin\angle Q} = \frac{|\overline{PT}|}{|\overline{PP^\prime}|} = \frac{|\overline{QT}| - |\overline{QP}|}{2a} = \frac{(a+c)-(a-c)}{2a}= \frac{2c}{2a}= \frac{c}{a} \tag{$\star\star$}$$

For the hyperbola, overlapping elements muddle the diagram a bit, but the argument is essentially the same (with a simple sign change):

enter image description here

$$\frac{\sin\angle S}{\sin\angle Q} = \frac{|\overline{PT}|}{|\overline{PP^\prime}|} = \frac{|\overline{QT}| + |\overline{QP}|}{2a} = \frac{(a+c)+(c-a)}{2a}= \frac{2c}{2a}= \frac{c}{a} \tag{$\star\star^\prime$}$$

There's no argument to make for the parabola, which has no "focal distance" or "major radius". However, one sees that, as the $\angle S$ nears $\angle Q$, the ratio of the lengths of these elements within an ellipse or hyperbola approaches $1$, as expected.


Thus, the "distance-to-focus-over-distance-to-directrix" ratio and the "focal-radius-over-major-radius" ratio (when defined) are the same constant that we happen to call the "eccentricity" of a conic. This discussion reveals the geometric meaning of that number. I suspect that most students these days had no idea that there is such meaning. Kudos to the teacher who assigned this problem as homework.

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  • $\begingroup$ This explains how to solve for how PF/PD = e. But it doesn't prove C/A = e. I know you mentioned that now I simply have to prove that C/A = e is equal to the same trigonometric ratio. But I need a proof that actually describes out the process of going from PF/PD = e to C/A = e I realize I keep rewording myself but this isn't something I fully understand so I am trying to make it as easy as possible. $\endgroup$
    – Dona
    Apr 27, 2012 at 0:04
  • $\begingroup$ @Dona: I don't want to give everything away in case this is homework. To prove $c/a = \sin S/\sin Q$, look at the figure "sideways", projecting into a plane perpendicular to the directrix. The cutting plane projects to a line; the cone --between apex (say, $A$) and cutting plane-- projects to a triangle (say, $\triangle ABC$); the Dandelin sphere projects to the incircle of that triangle. The ellipse's major axis projects to the "purple" segment ($BC$), and its center to the midpoint (say, $M$) of the segment. Use good ol' 2-dimensional trig to massage the ratio $MF/MB$ into $\sin S/\sin Q$. $\endgroup$
    – Blue
    Apr 27, 2012 at 1:07
  • $\begingroup$ This was a homework assignment for my son last week, and since I couldn't help him then I figured I would try and understand it, and find out what the answer was. Even though it's past, I beleive that knowing the answer would maybe allow me to help him in the future (even though he is far beyond anything I ever learned). He already knew how to solve PF/PD=e but not how to get c/a=sinS/sinQ (I guess). Thank-you for continuing to help me. $\endgroup$
    – Dona
    Apr 27, 2012 at 2:36
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    $\begingroup$ @sani: "We could have chosen any other point on the directrix. We still get to the result because the angle < S remains constant." ... Ah, but the angle doesn't remain constant! ... Consider a perfectly cubical room w/floor $\square ABCD$ & ceiling $\square A'B'C'D'$. Clearly, $\angle ABC$ is $90^\circ$; but $\angle AB'C=60^\circ$, because $\triangle AB'C$ is equilateral. As a pt $P$ travels from $B$ to $B'$ (and beyond), $\angle APC$ gets smaller. Only when the plane of $\triangle APC$ is perpendicular to $BB'$ do we get the largest possible angle. That's how "dihedral" angles are defined. $\endgroup$
    – Blue
    Mar 14, 2021 at 8:36
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    $\begingroup$ Thank you so much. I was struggling to understand this for 4 5 hours Thank you once again. $\endgroup$
    – anonymous
    Mar 14, 2021 at 10:41

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