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How would I prove that the distance from the focus F to any point P(x,y) on the ellipse equals the eccentricity times the distance from point P(x,y) to the vertical line x=a/e?

Fran
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Mar 05 '21 at 13:25
  • An ellipse has two focal points $F_1$ and $F_2$. Any point $P$ on the ellipse satisfies the property $|F_1 P|+|F_2 P|=2a$, where $|F_i P|$ is the distance from the focus $F_i$ to $P$ and $a$ is the larger radius. Maybe do some drawings and use the pythagorean theorem – Ahmed Hossam Mar 05 '21 at 13:25
  • @AhmedHossam: does this help answer the question ? –  Mar 05 '21 at 13:26
  • @Yves Daoust: maybe it helps to clear the confusion happening while thinking about "The focus" when discussing an ellipse? – Ahmed Hossam Mar 05 '21 at 13:30
  • @Yves Daoust: Outstanding experience in applied algorithmics and machine vision software development. Very nice! – Ahmed Hossam Mar 05 '21 at 13:36
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    @AhmedHossam: if you are interested, http://www.visionforvision.eu :-) –  Mar 05 '21 at 13:40
  • @Yves Daoust: yes, interested! – Ahmed Hossam Mar 05 '21 at 13:44

3 Answers3

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WLOG, the directrix is $x=0$ and the focus $(f,0)$. The given geometric constraint is (considering the squared distances)

$$(x-f)^2+y^2=e^2x^2.$$

By completing the square this is

$$(1-e^2)\left(x-\frac{f}{1-e^2}\right)^2+y^2=\frac{e^2f^2}{1-e^2}.$$

You recognize the equation of an ellipse, and the ratio of the axis is $\sqrt{1-e^2}$ (eccentricity $0$ for a circle).

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From the other direction:

$$\begin{align} (x-c)^2+y^2&=(x-c)^2+b^2\left(1-\frac{x^2}{a^2}\right)\\ &=x^2\left(1-\frac{b^2}{a^2}\right)-2cx+c^2+b^2\\ &=x^2\frac{c^2}{a^2}-2cx+c^2+b^2\\ &=\frac{c^2}{a^2}\left(x-\frac{a^2}c\right)^2-a^2+c^2+b^2\\ &=\frac{c^2}{a^2}\left(x-\frac{a^2}c\right)^2. \end{align}$$

user
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I found here a nice sketch, here it is:

enter image description here

The distance from a focus $S(ae,0)$ to any point $P(x_P,y_P)$ is $\sqrt{(ae-x_P)^2+y_P^2}.$ The eccentricity is $e$ and the distance from point $P(x_P,y_P)$ to the vertical line $x=a/e$ (directrix) is $ae - x_P.$ What you are trying to prove is:

$$\sqrt{(ae-x_P)^2+y_P^2} = e(ae-x_P)$$

Squaring both sides results in

$$y_P^2 = (e^2-1)(ae-x_P)^2$$

Further simplifying would result in the equation of the ellipse, if that's the case, if further simplifying leads to the equation of an ellipse, then the initial claim has to be true! Maybe try using $e = \sqrt{1-b^2/a^2}$?

Ahmed Hossam
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