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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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    Please provide enough context for us to understand what on earth you are talking about.2012-04-26
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    To me it's impossible to undertand your question. Could you explain what PF, PD, C, A, etc. mean?2012-04-26
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    It sounds like a conic section problem ... ?2012-04-26
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    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).2012-04-26
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    What is C? What is A?2012-04-26
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    I added to the OP what A and C are.2012-04-26

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