Derivation of the equation of Ellipse from its definition

    When we do the Kepler problem, we will obtain the result of a general conic section equation in polar form. But I got myself stuck looking at the equation of ellipse. I couldn't remember any damn thing I learned in my school, except the definition of ellipse. 

   But I know in mathematics, every thing can be derived from the fundamental postulates or axioms. So, I just proceed to build the equation starting from the definition. 

   The whole of the conic section is defined based on the concept of eccentricity. The definition I remember is that, Ellipse is the conic section with eccentricity value less than 1. And eccentricity is the ratio of the distance between focus to any point P to the perpendicular distance between point P and the directrix.

    To define a focus point, first we draw an axis and place the focus point. Directrix is the perpendicular line drawn anywhere to that axis except at the focus point.

   The condition is that you need to join all the points which obeys the rule that their eccentricity should be less than one and it is a constant. For each constant value, you will get different ellipse.    

   You don't even need the definition of axis to draw the ellipse, but it is just for convenience. All you need is just a point called focus and a line called directrix and follow the condition.
   

   From the definition of ellipse, we can see that there is two possible points which will obey the condition at the axis [which passes through the focus point and perpendicular to directrix]. Let me call those points $ P_1 and P_2 $ and the focus is F.  And also consider a general point P on the ellipse. 

    Giving separate symbols for each point helps to get clear understanding.  
   
    So far, we even don't know whether the ellipse would look symmetrical or not, or does have equal foci or etc. Yet, we haven't proved anything else.  

   The simple diagram we can draw with our so far concepts is,

From our definition, we know that the eccentricity value is constant for all the points of a given ellipse. Using the fact that   $P, P_1, P_2$ points are in the same ellipse, they all should have the eccentricity value - lets say "e".

    It gives $$ \frac{FP}{PM} = e and \frac{FP_1}{P_1M_1} = e and also, \frac{FP_2}{P_2M_1} = e $$

Let us say, $P_2P_1 = y $ 
   Using some elementary algebra, we can write $ \frac{FP_2}{P_2M_1} = e $ as, $$ FP_2 = e [P_2M_1] = e [P_2F + FP_1+P_1M_1]$$
   But $ FP_1 and P_1M_1$ are related by $$\frac{FP_1}{P_1M_1} = e $$ We can choose in either way.. I choose to write $P_1M_1$ in terms of $FP_1$ 
   So, $$ P_1M_1 = \frac{FP_1}{e}$$
Substituting this value in $FP_2$ we get,
  $$ FP_2 = e [FP_2 + FP_1 + \frac{FP_1}{e}]$$ 
$$ FP_2 = e \frac{[ e FP_2 + FP_1 (e+1)]}{e} $$
$$ FP_2 = e FP_2 + FP_1 (e+1) $$
$$ FP_2 = e (FP_2+FP_1) + FP_1$$ 
$FP_2 + FP_1 = P_2P_1 = y $
$$ FP_2 = e y + FP_1 $$ add $FP_1$ on both sides $$ FP_2 + FP_1 = e y + 2 FP_1 $$ and finally we get, $$ y = ey + 2FP_1$$
or simply, $$ FP_1 = \frac{y}{2} (1 - e) $$

That is all we want. If we draw our coordinate system at $\frac{y}{2}$ distance away from the focus on the axis, then focus would be situated exactly at $ \frac{y}{2}e $ distance away from the origin.

For simplicity, we take y = 2a so that y/2 = a. Thus, if the coordinate system is drawn at "ae" distance from the focus, the extreme points would have "2a" distance between them. And they are equidistant from the origin. 

Since we find out now that, both extreme points are symmetrical with each other with respect to the origin, we can conclude that based on symmetry arguments,
    Whatever the graph of ellipse drawn on the right side, would be similar if we choose the focus at the left side exactly at "-ae" distance from the origin at the axis of the ellipse. The same things would repeat. 

    Thus we conclude that, the graph of the ellipse has symmetry with respect to the y-axis - as shown in the figure below  

To find the distance of Directrix from the origin,

$$ OM_1 = OP_1 + P_1M_1 $$
where $OP_1 = a $ and $P_1M_1 = \frac{FP_1}{e} $ and $FP_1 = a - ae $
Gives, $$ OM_1 = a + \frac{a}{e} - a $$ 

Thus, we finally get the distance of directrix from origin as "a/e".

We are almost over finding the every relations along the axis of the ellipse. Now, we can go to the perpendicular to the axis i.e. the maximum height on the y - axis from the axis of the ellipse.

Since it deals with y - direction, let us consider the point $P_3$ which is the positive y - intercept of the ellipse. 

Again applying the eccentricity condition we get, $$ \frac{FP_3}{PM_3} = e $$
Drawing FT perpendicular to $P_3M_3$ helps us to use Pythagoras theorem, so that we can write everything in terms of known quantities. 
     
From Pythagoras theorem,
$$ {F_1P_3}^2 = {OP_3}^2 + {OF_1}^2 $$

Let us say, $ OP_3 = b $ for the sake of simplicity 
and we know $OF_1 = focal length = ae $ which gives 

So, $$ {F_1P_3}^2 = b^2 + (ae)^2 $$

From the geometry, it is understood that $TM_3 = F_1M_1$ and $ P_3M_3 = P_3T + TM_3$

$$ P_3M_3 = OM_1 = distance of the directrix from origin = \frac{a}{e}$$

And also $$ {F_1P_3}^2 = e^2 {P_3M_3}^2 = e^2 \frac{a^2}{e^2} = a^2 $$

Equating the two results of $ F_1P_3 $,
$$ b^2 + (ae)^2 = a^2 $$ Hence, 
$$b^2 = a^2 (1 - e^2) $$

Everything is in position to find the general equation. For the general equation, take a general point and apply the same old rules. 

$$ {FP}^2 = e^2 (PM)^2 $$ Let us take the co-ordinates of P as (x,y). 

Using the same Pythagoras theorem, distance between two points in a Cartesian plane can be determined using the formula,
If the points are $ A(x_1, y_1) and B(x_2, y_2)$ then distance between them $$ AB = \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Here we know the coordinates of all points i.e. F(ae,0), P(x,y), M(a/e, y). 
Using that , the equation $ {FP}^2 = e^2 {PM}^2 $ becomes, 
$$ (x - ae)^2 + y^2 = e^2 [ (x - \frac{a}{e})^2 + (y - y)^2] $$
$$ (x - ae)^2 + y^2 = (xe - a)^2 $$
$$ x^2 + (ae)^2 -2xae + y^2 = (xe)^2 + a^2 -2xae $$
$$ x^2 - (xe)^2 +y^2 = a^2 - (ae)^2 $$
$$ (1-e^2) x^2 + y^2 = a^2 (1- e^2) $$
Diving the equation by $ a^2 (1- e^2) $ We get,

$$\frac{x^2}{a^2} + \frac{y^2}{a^2(1 - e^2)} = 1 $$
But we know b^2 = a^2 (1 - e^2)

Hence, we derived the general equation of Ellipse in Cartesian co-ordinates as,

$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$ where a > b

Properties (measured from origin):
Focal length = ae
Distance of the directrix = a/e
$ b^2 = a^2 (1 - e^2) $
Length of the major axis = 2a
Length of the minor axis = 2b  

Note: Why Ellipse are important?

We know that from Kepler's laws of planetary motion, planets follow Elliptical path with the Sun at one of its two foci . 
    Of course, when we solve the equation of motion for the planets using Newton's law of Gravitation, Elliptical path is obtained as one of the results of the general conic section. 
   But it was a great discovery at that time, because there was no any Gravitational law or anything to explain the motion of the planets. All they got is data.
   Kepler predicted this path, just from observing the loads of data from the danish astronomer Tycho Brahe. All these things are discovered just because they have a profound knowledge in Conic Section and their equations.