COMMON VERTEX EQUATION

Common Vertex Equation

    There is a way to write a single equation that describes all conic sections. To do this we need the 'vertex equation' for each conic. This equation has the vertex of the conic at the origin.

    As we progress through these equations we will discover a deeper connection between the conics; this connection involves the eccentricities of the conics.

                    Circle ε = 0

                    Parabola ε = 1

                    Ellipse 0 < ε < 1

                    Hyperbola 1 < ε

    Let's begin with the parabola. It's vertex equation is:

                    y²= 2px

    This equation is slightly different from the one we've been using all along. It's more convenient now to use the constant 2p instead of 4c. The values of each must be equal so c = ½ p. Recall that c is the distance from the vertex to the focus and from the vertex to the directrix; p is, therefore, the distance from the directrix to the focus. You can see this in the graph below.

In summary, the parameters for the conics are:

We will use these parameters next.

Before we proceed, let's take another look at these equations and why the conics are named the way they are.

Now, if we introduce the eccentricity ε = c/a

for an ellipse:

for a hyperbola:

Substituting this in for an ellipse we get:

and for a hyperbola we get

   y² = 2px  + (ε² − 1)x²   = 2px − (1 − ε²)x²

The equations are the same.  This equation is called the

common vertex equation.

         y²  = 2px − (1 − ε²)x²

If ε = 1, then (1 - ε²) = 0, and we have a parabola.

For the other cases the standard form of this equation looks like:

        y² + (1 - ε²)x² − 2px = 0  

The + sign in front of (1 - ε²) is important.

    For an ellipse, 0 < ε < 1  so   (1 - ε²) > 0 and the + sign remains.

    For a hyperbola,  1 < ε,  so (1 - ε²) < 0 and the + sign becomes a − sign.

    For a circle,   ε = 0, then (1 - ε²) = (1 - 0) = 1, and the  + sign remains.

The following graph shows this relationship between the conic sections.

Now let's see how this works with the equation.

Now, if ε = 0 then this equation becomes:

        y²= 2px − (1- 0) x²= 2px − x²

Simplify.

        y²= 2px − x²

Bring all terms to one side and leave 0 on the other side.

        y² + x² − 2px = 0

Now complete the square.

        y² + x² − 2px + p²= p²

Factor

        y²+ (x - p)²= p²               This is a circle centered at (p,0) with radius p.

Let ε = 1 then this equation becomes

substituting 1 in for x

        y²= 2px - (1 - 1)x²= 2px - (0) x²

Simplify.

            y²= 2px                   This is a parabola with vertex at (0,0).

Let 0 < ε < 1

the equation becomes:

When ε < 1 then (1 - ε²) is greater than 0 so we have

            y²= 2px − (1 − ε²)x²       (no change, the minus sign after 2px remains)

to simplify, let k = (1 − ε²)      (from our observations k > 0)

Substitute k into the equation.

            y²= 2px − kx²

Bring all terms to one side and leave 0 on the other side.

               y² + kx² − 2px = 0

Now complete the square. (This is why we created k.)

Factor.

            .

let m = p/k ( so k = p/m) we do this to simplify the algebra

    We have an ellipse (because of the + sign and because the denominators are not equal (unless m=p which implies that ε = 0 and we would have a circle)  and whose center is at (m, 0) and whose major axis is along the X axis and whose minor axis is parallel to the Y axis, and here's why:

            a² = m²  and  b²= mp

recall m = p/k and k = (1 −  ε²)

            a²= m²= (p/(1 −  ε²))²  = p²/(1 −  ε²)²

            b²= mp = (p/(1 −  ε²))p = p²/(1 −  ε²)

    Now, in this case, (1 − ε²) > 0, and is also less than 1. Now if you square a number that is less than one, its square is even smaller. For example (1/2)²= 1/4 and if you divide the same number by a smaller number that number becomes larger, for example 12/6 = 2 but 12/4 = 3. It follows then a² > b². We have an ellipse whose major axis is along the X axis and whose minor axis is parallel to the Y axis. This makes sense since we started with an ellipse in its vertex position.

Let ε > 1

First, if ε > 1 then (1 − ε²) is less than 0 so we have

let k = (1 −  ε²) from our observations k < 0

Substitute in.

            y²= 2px −  (1− ε²)x²

            y²= 2px + kx²         (k < 0 so −k  =  +k)

            y² − kx² − 2px = 0

Now complete the square.  (first factor out the − sign: note the parenthesis)

Factor.

            .

let m = p/k ( so k = p/m)    (we do this to simplify the algebra, again)

The steps are the same as for the ellipse previously, except we divide by a −m².

and we have a hyperbola (− sign in front of the y² term) whose center is at (-m, 0)

now:

a² = m² and b² = mp

recall m = p/k and k = (1 −  ε²)

so mp = p²/|(1 −  ε²)| and m² = p²/(1 −  ε²)²

a² = m² = p²/(1 −  ε²)²

|b²|= |mp| = |p²/(1 −  ε²)|

    And note that the negative sign is in front of the y² term which means this hyperbola opens along the X axis where

|a| > |b| when 1 < ε < √2 ,

|a| = |b| when ε = √2,

|a| < |b| when ε > √2.

reasoning:

when ε = √2, then ε² = 2 and (1 −  ε²) = 1 – 2 = -1

    |b²|= |p²/(-1)| = p²

and (1 −  ε²)² = (1 – 2)² = 1

       a² = |p²/(1) | = p²

so |a| = |b|

when ε < √2 (but > 1) then ε² < 2 and |(1 −  ε²)| < 1 which

means

|(1 −  ε²)| > (1 - ε²)²

             1 > (1 - ε²)² / |(1 −  ε²)|

now:

|a²| / |b²|

(p²/ |(1 - ε²)|) / (p² / (1 - ε²)²) = (1 - ε²)²  / |(1 - ε²)|

a² > b²

so |a| > |b|

when ε > √2 then ε² > 2 and |(1 - ε²)| > 1 which means

|(1 - ε²)| < (1 - ε²)² so

p²/ |(1 - ε²)| > p² / (1 - ε²)²       (another approach, divide by a smaller number)

b² > a²

so |a| < |b|