Confusion with the eccentricity of ellipse

Question

Confusion with the eccentricity of ellipse. On wikipedia I got the following in the directrix section of ellipse.

Each focus F of the ellipse is associated with a line parallel to the minor axis called a directrix. Refer to the illustration on the right, in which the ellipse is centered at the origin. The distance from any point P on the ellipse to the focus F is a constant fraction of that point's perpendicular distance to the directrix, resulting in the equality e = PF/PD. The ratio of these two distances is the eccentricity of the ellipse. This property (which can be proved using the Dandelin spheres) can be taken as another definition of the ellipse. Besides the well-known ratio e = f/a, where f is the distance from the center to the focus and a is the distance from the center to the farthest vertices (most sharply curved points of the ellipse), it is also true that e = a/d, where d is the distance from the center to the directrix.

It is given that $e=\frac fa=\frac ad$

In my book it was only given that $e=f/a$ (in my book there is nothing given about directrix of an ellipse).

My question

Knowing that $e=f/a$ how can I get $e=a/d$ and $e=PF/PD$?

Answer 1

Hint: If the eccentricity $e$ & the major axis $2a$ of an ellipse are known then we have the following

1. Distance of each focus from the center of ellipse $$=\text{(semi-major axis)}\times \text{(eccentricity of ellipse)}=\color{red}{ae}$$
2. Distance of each directrix from the center of ellipse $$=\frac{\text{semi-major axis} }{\text{eccentricity of ellipse}}=\color{red}{\frac ae}$$

Answer 2

If I understood your question correctly, you're essentially asking how one can find the equation for the directrix if one only has the equation for an ellipse with a given eccentricity.

You start with the equation below $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \qquad(1) $$ where $a$ and $b$ are positive real-valued constants.

If you then define two points on the $x$ axis, $F$ and $F'$, by their coordinates, $$ F \equiv (\epsilon a, 0) \qquad(2) $$ $$ F' \equiv (-\epsilon a, 0) \qquad(3) $$ for some as-yet undetermined real value $\epsilon \ge 0$, and then require that the sum of the distances from $F$ to $P$ and from $P$ to $F'$ be equal to $2a$ - for any arbitrary point $P$ on the ellipse - you'll find out that that is only possible if you choose $\epsilon$ to satisfy $$ \epsilon = \sqrt{1 - \frac{b^2}{a^2}} \qquad(4) $$

So far, the above has nothing to do with the directrix. It's just a way to construct the foci and find the eccentricity, starting from some equation and then requiring that the curve described by that equation must have the fundamental property of an ellipse (namely, that the sum of the distances from any point on the ellipse to the two foci is a constant).

Now, let's see how we can prove the directrix property. Imagine that there is a vertical line at $x=d$ for some as-yet-undetermined real value $d \ge 0$ and let's see if we can find a value of $d$ such that the directrix property is satisfied. The focus for positive $x$ is $F$ (see its coordinates above), and an arbitrary point $D$ in the would-be directrix has coordinates $D = (d, y)$. The directrix property mandates that $$ \epsilon = \frac{\overline{PF}}{\overline{PD}} \qquad(5) $$

Now, let's look at the square of the distances involved, since that gets rid of the square roots: $$ \overline{PF}^{\,2} = (x - \epsilon a)^2 + (y - 0)^2 = (x - \epsilon a)^2 + y^2 $$

But $P$ lies on the ellipse so $(x,y)$ satisfies $(1)$. Therefore, eliminating $x^2$ in favour of $y^2$, but leaving the $x$ term alone, we have $$ \overline{PF}^{\,2} = a^2\,(1 + \epsilon^2) + y^2\,(1 - \frac{a^2}{b^2}) - 2\epsilon ax $$

How about $\overline{PD}^{\,2}$? Note that $D$ and $P$ have the same $y$ coordinate so $$ \overline{PD}^{\,2} = (x - d)^2 + (0)^2 = (x - d\,)^2 $$

Here's the crux now. Can we find a value of $d$ such that $(5)$ is true? Imposing $$ \frac{\overline{PF}^{\,2}}{\overline{PD}^{\,2}} = \frac{a^2\,(1 + \epsilon^2) + y^2\,(1 - \frac{a^2}{b^2}) - 2\epsilon ax}{(x - d\,)^2} = \epsilon^2 $$ we get $$ a^2\,(1 + \epsilon^2) + y^2\,(1 - \frac{a^2}{b^2}) - 2\epsilon ax = \epsilon^2\,(x - d\,)^2 $$

Note that the complicated expression above is something like this: $$ (\mbox{terms independent of $x$ and $y$}) + (\mbox{terms involving $y^2$}) - 2\epsilon ax + 2\epsilon^2 xd = 0 $$ because the $x^2$ terms can be removed using $(1)$. Since the above has to be true for all $x$, the only hope for $(5)$ to be possible is if we choose $d$ such that $$ 2\epsilon ax - 2\epsilon^2 xd = 0 \qquad\Rightarrow\qquad d = \frac{a}{\epsilon} $$

With this choice, it's then not hard to show that the terms not written above also vanish so, indeed, $$ \epsilon = \frac{\overline{PF}}{\overline{PD}} $$ is true, provided we make that choice for $d$.

So, what's the conclusion? The conclusion is that an ellipse has two equivalent properties, namely, (A) the sum of the distances from the foci to any arbitrary point on the ellipse is a constant, and (B) the ratio of (the distance from an arbitrary point on the ellipse to one of the foci) to (the distance from that point to a fixed vertical line parallel to the semi-minor axis) is equal to the eccentricity of the ellipse.

And we also proved that $\epsilon = a/d$.

(Technically, I only proved that (A) above implies (B). We'd also have to prove that (B) implies (A) for the equivalence I referred to be proven. That's an exercise I leave for the reader...)

Answer 3

The ellipse can also be defined as the locus of points whose distance from the focus is proportional to the horizontal distance from a vertical line known as the conic section directrix, where the ratio is $<1$. Letting $e$ be the ratio and $d$ the distance from the center at which the directrix lies, then in order for this to be true, it must hold at the extremes of the major and minor axes, so

$r=\frac{a-c}{d-a}=\frac{\sqrt{b^2+c^2}}{d}$

With a bit of solving, you get

$d= \frac{a^2}{\sqrt{a^2-b^2}} = \frac{a^2}{c}$ and

$e = \frac{\sqrt{a^2-b^2}}{a}=\frac{c}{a}$.

From the value of $d = \frac{a^2}{c}$

or $\frac{d}{a} = \frac{a}{c} = \frac{1}{e}$

Hence $e = \frac{a}{d}$

In this proof, you can see that we have derived d and e from the extremities, since their values are easily available. For the $\frac{PF}{PD}$ ratio, you can put $PF = a - c$ and $PD = d - a$ (similar to what we did for finding d and e).

Hope this helped. Cheers!