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.

x

^{2}

**/**a

^{2}+ y

^{2}

^{ }

**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:

x

^{2}**/**16 + y^{2}^{ }^{}^{ }**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 =

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 =

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

This means the figure is a perfect circle. For a circle the eccentricity e = 0. For our own example here: e =

**Ö**8**/**4 = 2.82/ 4 = 0.705When, 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:

x

^{2}**/**8**+ y**^{2}^{ }^{}^{ }**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:

x

^{2}**/**16 + y^{2}^{ }^{}^{ }**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 x

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

9(x

Complete he square for each term in parenthesis:

9 (x

Or:

9 (x

Divide through by 36:

(x + 2)

To

Then write:

x'

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

We also find:

c =

So: c =

So the foci are situated at the points (0, 9 x

^{2}**+ 4 y**^{2}^{ }^{}^{ }+ 36 x - 8 y + 4 = 0To get this into a form that can easily be inspected and graphed, first re-arrange the terms to get:

9(x

^{2}**+ 4 x) + 4 (y**^{2}^{ }^{}^{ }- 2 y) = - 4Complete he square for each term in parenthesis:

9 (x

^{2}**+ 4 x + 4) + 4 (y**^{2}^{ }^{}^{ }- 2 y + 1) = - 4 + 4 + 36Or:

9 (x

^{2}**+ 4 x + 4) + 4 (y**^{2}^{ }^{}^{ }- 2 y + 1) = 36Divide 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 }= 4So: c =

**Ö**(9 - 4 ) =**Ö**5_{}^{}__+__

**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 x

^{2}

**+ 9 y**

^{2}

^{ }

^{}^{ }= 144

b) 4 x

^{2}

**+ y**

^{2}

^{ }

^{}^{ }= 1

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

x

^{2}

**/7**+ y

^{2}

^{ }

^{}^{ }

**16 = 1**

^{ }/And sketch the graph showing the locations of the foci.

## No comments:

Post a Comment