derivation of ellipse parameters

Question

At various places on the Web (including Mathematics StackExchange) are various methods of calculating the semimajor and semiminor parameters of a ellipse $(a, b)$ from the location of a focus on the $x$ axis at a specified distance $c$ from the origin and a perpendicular from a directrix along the $x$ axis. However, I have not found anywhere where these can be derived in closed form from only the information available in the focus-directrix definition of an ellipse, i.e., the locus of points, the ratio of whose distance from the focus point and a perpendicular from the directrix is a constant, $0<e<1$. The derivations I have found either explicitly state values for the location of the focus along the $x$ axis $(ae)$ and the equation for the directrix $(x=a/e)$ and/or presuppose the existence of a second focus equidistant from the origin. Neither of these conditions should be necessary for derivation from the definition alone.

To this end, consider the folowing diagram:

$F$ is the focus, at distance $c$ from the origin. $DD'$ is the directrix, at distance $d$ from the origin. Eccentricity is specified as $e$. This is all the information we have from the definition.

Let $A$ be a point on the ellipse which is also on the $x$ axis. Let $B$ be a point on the ellipse which is also on the $y$ axis. Find $a = f(c,d,e)$ and $b=f(c,d,e)$

$\frac{a-c}{d-a} = e $ by definition applied to the $x$ axis, so $de = (1+e)a-c$

$de = \sqrt{b^2+c^2}$ by definition applied to the $y$ axis.

So $(de)^2 = b^2+c^2= ((1+e)a-c)^2 = \\ =(1+e)^2a^2-2(1+e)ac+c^2 \\ b^2 = a^2+2ea^2+e^2a^2-2ac-2eac$

Also,

$c = (1+e)a-de \\ c^2=((1+e)a-de)^2 = (1+e)^2a^2-2(1+e)de + (de)^2 \\ = a^2+2ea^2+ e^2a^2-2de-2de^2+d^2e^2$

$(de)^2 = b^2+c^2 = a^2+2ea^2+e^2a^2-2ac-2eac + a^2+2ea^2+ e^2a^2-2de-2de^2+(de)^2 \\ 2de(1+e)= 2a^2 + 4ea^2 + 2e^2a^2 - 2ac - 2eac\\ de(1+e) = ((1+e)a)^2 -2(1+e)ac\\$

or

$de = (1+e)a^2 - 2ac\\$

Completing the square:

$de(1+e)+c^2 = ((1+e)a - c)^2$

If I'm correctly interpreting everything that has been known about ellipses for the past 2000 years or so, I should be able to reduce this to $a=de$, which obviously can't be done without some substitution for $c$. Any assistance is greatly appreciated.

Answer 1

$c$,$d$ and $e$ are not all independent variables; the value of 2 of them determines the third. The relationships are $c=de^2$, $d=\frac{c}{e^2}$ or $e=\sqrt{\frac{c}{d}}$. These all can be derived by substituting either of the accepted parameters $d=\frac{a}{e}$ or $c=ae$ into $\frac{a-c}{d-a}=e$. An expression for $a$ follows directly from this equation. An expression for $b$ comes from $\sqrt{b^2+c^2} = de =d\sqrt{\frac{c}{d}}$.

$e=\sqrt{1-\frac{b^2}{a^2}}$ can be derived directly from the latter equation by simple algebra. Proof of the general equation for a standard ellipse can then proceed as can be found in various sources.

Answer 2

Your $y$ axis is placed at a distance $d$ from the directrix, but that distance is not arbitrary. If you take $x$ axis as the ellipse major axis, then the ellipse intersects it twice: at points $A=\big({c+ed\over1+e},0\big)$ and $B=\big({c-ed\over1-e},0\big)$. The center $O$ of the ellipse is the midpoint of $AB$ and we have then: $$ {c+ed\over1+e}+{c-ed\over1-e}=0, \quad\text{that is:}\quad d={c\over e^2}. $$ Substituting that into the expression for $A$ gives then $a=c/e$ and finally $de=c/e=a$. That immediately yields $a=\sqrt{b^2+c^2}$.

EDIT.

Let's show that the points on the ellipse with maximum distance from the major axis lie on the vertical bisector of $AB$. If $P=(x,y)$ is on the ellipse, we have by definition: $$ {\sqrt{(c-x)^2+y^2}\over d-x}=e. $$ Squaring and rearranging leads to: $$ y^2=e^2(d-x)^2-(c-x)^2=(e^2-1)x^2-2(e^2d-c)x+e^2d^2-c^2, $$ which reaches its maximum value for $$ x={c-e^2d\over 1-e^2}={x_A+x_B\over2}. $$

Answer 3

Here's a formal proof incorporating Arentino's suggestions:

An ellipse can also be defined as the locus of points, the ratio of whose distances between a point (the focus) and the perpendicular to a line (the directrix) is a constant, $0 < \mathbf{e} <1$.

Given directrix $\mathbf{XX'}$ and focus $\mathbf{C}$:

Draw a line through $\mathbf{C}$ perpendicular to $\mathbf{XX'}$ meeting $\mathbf{XX'}$ at $\mathbf{D}$ (Euclid Pr. I.12).

Without loss of generality, regard $\mathbf{CD}$ as the horizontal axis of a Cartesian coordinate system.

Let $\mathbf{A}$ be the point on this axis between $\mathbf{C}$ and $\mathbf{D}$ satisfying Definition.

Then $\mathbf{|AC|} = e \cdot \mathbf{|AD|}$

Let $\mathbf{A'}$ be another point on this axis but not between $\mathbf{C}$ and $\mathbf{D}$ satisfying Definition.
Since $\mathbf{|A'C|} < \mathbf{|A'D|}$, this point must be to the left of $\mathbf{C}$ in the Figure.

Draw perpendicular bisector to $\mathbf{AA'}$, $\mathbf{YY'}$, intersecting $\mathbf{AA'}$ at origin $\mathbf{O}$ (Euclid Pr. I.10)

Let $\mathbf{|A'O|} = \mathbf{|OA|} = a$

Let $\mathbf{|OD|} = d \;$. Then $\mathbf{|AD|} = d-a, \; \mathbf{|A'D|} = d+a$

Let $\mathbf{|OC|} = c \;$. Then $\mathbf{|AC|} = a-c, \; \mathbf{|A'C|} = a+c$

Then, from Definition:

$$\frac{a-c}{d-a} = e\\ \frac{a+c}{a+d} = e\\ a-c = de-ae \\ a+c = de+ae\\ 2a = 2de\\ a=de$$

Now : $$\frac{a+c}{a+\frac{a}{e}}=e\\ a+c = 1 \cdot ae+a \\ c = ae$$

Let $\mathbf{B}$ be a point on $\mathbf{YY'}$ satisfying Definition. Draw $\mathbf{BC}$, creating right triangle $\mathbf{BOC}$

$b^2 + c^2 = \mathbf{|BC|}^2\\$ (Euclid Pr. I.47).

But also, from Definition, $ \;\mathbf{|BC|}^2 =(de)^2\\$

And $(de)^2=a^2$ as shown above

So $a^2=b^2+c^2 = b^2 + (ae)^2$.
Simplifying, $e = \sqrt{1-\frac{b^2}{a^2}}$

Now locate point $\mathbf{C'}$ at distance $c$ from $\mathbf{O}$.

$\mathbf{\triangle BOC \cong \triangle BOC'}$ (Euclid Pr. I.4), so $\mathbf{|OC'|} =a \\$

And the ellipse thus constructed has identical properties to that constructed by specifying $\mathbf{C}$, $\mathbf{C'}$ and $a$

From here we can derive the ellipse standard form equation using $c=ae, \; a=de$.

For an arbitrary point $\mathbf{P}(x,y)$ on the ellipse, $x>c$:

From Definition: $e \cdot (d-x) = a - ex = \sqrt{(x-c)^2 + y^2}$ $$a^2-2aex + e^2x^2 = x^2-2cx+c^2+y^2 = x^2-2aex+c^2+y^2$$ $$a^2 + (1-\frac{b^2}{a^2})x^2 = x^2 + c^2 + y^2$$ $$a^2-c^2 = b^2 = \frac{b^2}{a^2}x^2 + y^2$$ $$1 = \frac{x^2}{a^2} + \frac{y^2}{b^2}$$ For $x<c$, use $d+x, \;x+c$ to arrive at the same equation