## Saturday, January 14, 2017

### Looking At Analytic Geometry (2): The Ellipse

The ellipse is a geometric figure - curve we examined before in the context of real planet orbits. For example, in the diagram shown above, a planet at point P(x,y) in its orbit is referenced to the two foci of the ellipse at f1 and f2. As the point P moves along the total connecting distance f1-P-f2 remains constant. By inspection what we call the semi-major axis is a = 4 and the semi-minor axis is approximately 2.8.

If we know the general form of the equation for an ellipse, e.g.

x2 / a 2 +    y2  / b 2 = 1

We could make a fair guess at what the actual equation of the orbit shown would be, say taking

b =   Ö 8  =   2.82

The ellipse equation would then be:

x2 /16 +    y2   / 8 = 1

And the graph would be plotted as shown below:

Which is a solid match for the earlier sketch orbit.

What is c, mathematically?

It is the square root of the difference between the semi-major axis squared, and the semi-minor axis squared:

c =   Ö (a 2  -   b 2)

In this case, we'd get:    Ö (16  -  8) =   Ö 8  =   2.82

That means the coordinates of the foci are f1 (- Ö 8, 0)  and f2 (0,  - Ö 8)

Note that when c = 0 we have a = b. (One constant fixed radius: e.g. r = a = b)

This means the figure is a perfect circle. For a circle the eccentricity e = 0. For our own example here: e =  Ö/ 4   =   2.82/ 4  = 0.705

When, however, a > b then the shape alters to elliptical, but also when b > a.

If we juxtapose the values for a, b in the general equation we get:

x2 / 8  +    y2   / 16 = 1

And the ellipse orientation changes to that shown on the graph below:

From the preceding it is possible to sketch the graph of a given ellipse, as well as give its eccentricity e (= c/a) from just the coordinates of the eclipse center,. the focus F (at least one) and the semi-major axis a.

Example problem: Sketch the graph and find e for the ellipse with center coordinates C(0,0), focus coordinate F(0, 2) and a = 4

Solution: Given the coordinate F(0,2) we must have a > b   and hence, c = 2. Then the semi-minor axis is found from:

b =   Ö (a 2  -   c 2)    =   Ö (16  -   4 )    =  Ö 12

The eccentricity e = c / a    = 2/ 4   = 0.5

The equation for this ellipse will be:

x2 /16 +    y2   / 12 = 1

Which figure is graphed below:

As in the case of the circle, ellipses aren't always centered at (0,0) or the origin of an x-y grid. Consider the equation below:

9 x2  +   4  y2   + 36 x   - 8 y + 4 = 0

To get this into a form that can easily be inspected and graphed, first re-arrange the terms to get:

9(x2  +   4 x)  +  4 (y2   - 2 y)  =  - 4

Complete he square for each term in parenthesis:

9 (x2  +   4 x  +  4)  +  4 (y2   - 2 y   + 1)  =  - 4   + 4   + 36

Or:

9 (x2  +   4 x  +  4)  +  4 (y2   - 2 y   + 1)  =  36

Divide through by 36:

(x + 2) 2 / 4 +    (y - 1 ) 2   / 9 = 1

To simplify further, let x' = (x + 2) and y' = ( y - 1)

Then write:

x' 2 / 4 +    y ' 2   / 9 = 1

Which yields the ellipse shown below once the x, y- coordinate offsets are taken into account:

We also find:

c =   Ö (a 2  -   b 2)     where in this case,  a 2  =  9   and   b 2  =   4

So:  c =   Ö (9   -   4 )    =   Ö 5

So the foci are situated at the points (0,  + Ö 5 ) in the new coordinate system or at:

(-2,  1 +   Ö 5) in the original. Which values can be checked from the graph.

Problems:

1) In the last example problem worked above, we saw that the ellipse equation of form:

(x + 2) 2 / 4 +    (y - 1 ) 2   / 9 = 1

Could be simplified to: x' 2 / 4 +    y ' 2   / 9 = 1

Using:   x' = (x + 2) and y' = ( y - 1)

Show how the offset affects the location of the simplified ellipse form by graphing both ellipses on the same axes. Thus, satisfy yourself that the graph shown is justified once the x, y- coordinate offsets are taken into account.

2) An ellipse has center coordinates at C(-3,0) and focus at F(-3, -2) with a = 4. Find the eccentricity e and sketch the graph.

3) An ellipse has center coordinates at C(2, 2) and focus at F(-1, 2) with a = Ö 10. Find the eccentricity e and sketch the graph.

4) Sketch the following ellipses from their equations below, and give the semi-major axes, semi-minor axis and eccentricity e.

a) 4 x2  +   9  y2   = 144

b) 4 x2  +    y2   =    1

5) Look up the definition of the "directrix" of an ellipse. Then find the eccentricity and directrices of the ellipse:

x2 /7 +    y2   / 16 = 1

And sketch the graph showing the locations of the foci.