First, some housekeeping for Maple:

>

restart:

>	with(plots):

Warning, the name changecoords has been redefined

You will aso see below some things like "scaling=constrained" and "thickness=3" which tell Maple how I want the pictures to look: They don't change the math, you can ignore them!

We are going to look at conic sections, and we have the eccentricity as a way of telling them apart.

If I choose an eccentricity between 0 and 1 for an example, we will get an ellipse:

>

e := 0.6;

$e:=.6$

If we remember the relations between a, b, and c, for an ellipse in standard position with its foci on the x-axis, e = c/a and b^2 = a^2 - c^2, so we could pick a value for a to go with that eccentricity and then know what the ellipse looked like.

I'll pick a = 5, so the ellipse will cross the x-axis at +/-5:

>

a := 5;

$a:=5$

>	c := a * e; b := sqrt(a^2-c^2);

$c:=3.0$

$b:=4.000000000$

Then the ellipse has to be:

>	implicitplot(x^2/a^2+y^2/b^2=1,x=-a..a,y=-b..b, scaling=constrained, thickness=3);

That looks right: The major axis is 2a = 10 units long, and the minor axis is 2b=8 units.

The foci are at +/- c on the x-axis. The directrices are the vertical lines x = +/- k, where k = a^2/c. I am going to use the left-side directrix below so I will set

>	k := -a^2/c;

$k:=-8.333333333$

If we plot the ellipse along with the directrices we see

>	implicitplot([x^2/a^2+y^2/b^2=1,x=k,x=-k],x=-a+k..a-k,y=-b..b, scaling=constrained, thickness=3);

Since we have not been paying attention to directrices except for parabolas, note these lines DO work: For any point on the ellipse, the distance to a focus is the eccentricity times the distance to a directrix. For example, at the "top", the point (0,4), the distance to either focus is 5 and the distance to a directrix is 25/3, and (3/5) times (25/3) does give 5.

In fact we can use that to derive the equation k = a^2/c above: Pick one of the ends of the major axis, say (a,0) in general. The focus on that side is at (c,0). If the vertical like x=k is the asymptote on that side: The distance from (a,0) to (c,0) is a-c. The distance from (a,0) to x=k is k-a. So if the eccentricity is right, a-c = e ( k-a). If you put in e=c/a and solve for k you get a^2/c.

Now on to polar coordinates:

Equation (2) on page 734 gives us an equation in polar coordinates for any conic. But there is a misleading statement. The book says "Where x=k>0 is the vertical directrix", but that turns out to flip the picture left-to-right. We will use the left directrix, where k<0, as mentioned above.

Notice a nice thing about doing this in polar coordinates: we don't have to remember different equations for ellipse, parabola, and hyperbola since the eccentricity will tell them apart.

But the equation is for a conic with a focus at the origin, not the center at the origin, so let's shift our rectangular-coordinates ellipse c units to the right to put a focus at (0,0), so it should match what we get in polars:

>	implicitplot([(x-c)^2/a^2+y^2/b^2=1,x=-k+c,x=k+c], x=-a+c+k..a+c-k, y=-b..b, scaling=constrained, thickness=3);

So if we use the polar equation with the eccentricity and directrix values above we should see the same ellipse! But we need to change k to go with the shifted axes:

>

k := k+c;

$k:=-5.333333333$

The equation is then

>	R = (K * E)/(1 + E*cos(theta));

$R=\frac{KE}{1+E\cos \left(\theta \right)}$

>	plot(k*e/(1+e*cos(theta)), theta=0..2*Pi, coords=polar, scaling=constrained, thickness=3);

And you can see we have the same ellipse!

I won't go back through ALL of it for hyperbolas and parabolas, but here is the result of changing just e:

>

e := 1.5;

$e:=1.5$

>	plot(k*e/(1+e*cos(theta)), theta=0..2*Pi, coords=polar, scaling=constrained, thickness=3);

THAT DOES NOT LOOK MUCH LIKE A HYPERBOLA! But remember that a hyperbola goes off to infinity: Maple tried to scale the picture so we could see more of it. If we limit the picture to stay near the origin,

>	plot([k*e/(1+e*cos(theta)),theta, theta=0..2*Pi], -30..20, -20..20, coords=polar, thickness=3);

Note that Maple gets confused when the r values "go to infinity", that's where it draws what are actually the asumptotes.

Or we can plot a parabola with the same equation, by setting the eccentricity to 1:

>

e := 1;

$e:=1$

>	plot([k*e/(1+e*cos(theta)),theta, theta=0..2*Pi], -10..20, -20..20, coords=polar, thickness=3);

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