Distances are equal in an ellipse.

Question

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?

Answer 1

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).

Answer 2

From the other direction:

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

Answer 3

I found here a nice sketch, here it is:

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 $e = \sqrt{1-b^2/a^2}$?