Welcome to a second video
on polar equations of conic sections.
In this video we'll take a look at graphing an ellipse.
But let's start with a quick review.
If we have a polar equation in either of these two forms,
we'll have a conic section with eccentricity E.
Then if it's in this form with cosine theta,
we have a vertical directrix at X equals or minus D.
And if the equation has sine theta in it,
we'll have a horizontal directrix Y equals plus or minus D.
For more details about this thing,
we want to take a look at part one.
But in this video,
we're going to take a look at another example.
Here we have R equals eight,
divided by the quantity four plus two sine theta.
Our first step is to make sure our denominator
is in the form of one plus or minus E sine theta.
So we want this four to be a one,
so we'll divide everything on the right side
by positive four.
Let's go ahead and simplify this.
So we'll have two divided by one plus 1/2 sine theta.
Okay, so in this form,
we can gather some important information.
The first thing we should notice
is the coefficient of sine theta is equal to 1/2.
So eccentricity is equal to 1/2, so we have an ellipse.
Notice the numerator is E times D.
So E times D is equal to positive two.
But we know E is equal to 1/2,
so we have 1/2 times D is equal to two.
Well that tells us that D is equal to four.
And since the equation contains plus E sine theta,
we have a horizontal directrix at Y equals positive four.
Let's go ahead and take this information
onto the next screen and start to graph our ellipse.
Let's start by graphing our directrix at Y equals four.
It would be somewhere in here.
Now, I've mentioned in the previous video
that remember on the unit circle,
sine theta is equal to Y.
That helps me remember that
since this equation contains sine theta,
the major axis of this ellipse
will be along the Y axis here.
So the ellipse will look something like this,
where one focus is at the pole.
And the reason I find that helpful is
we can find the endpoints of the major axis
by selecting theta equal to pi over two,
and three pi over two.
So let's go ahead and do that.
Now we can use either the original form
or the simplified form to determine the radius.
Let's go ahead and just use the original form.
So when theta is pi over two have sine of pi over two
which is equal to one,
so R is going to equal eight divided by four plus two,
or eight divided by six,
which is equal to 4/3.
And then when theta is three pi over two,
the sine of three pi over two
is going to equal negative one,
so we'll have eight divided by four plus negative two
or eight divided by two,
which is equal to four.
Let's go ahead and plot these two points,
which are the endpoints of the major axis.
So we have pi over two 4/3
somewhere in here.
And three pi over two four
would be here.
Again, these are the endpoints of the major axis.
And don't forget, we have a focus at the pole.
Let's go ahead and pick theta equal to zero radians
and also pi radians.
When theta is equal to zero,
since sine zero is zero,
we have eight divided by four, which is two.
That would be this point here.
And then the sine of pi would also be zero,
so we'd have two again.
So it would be right here.
Remember, the center of this ellipse
would be the midpoint of this major axis.
Since the major axis is four units plus 4/3 units long,
we can take this entire length and divide by two
to find the center of the major axis,
to find the center of this ellipse.
Let's go ahead and do that.
Four plus 4/3 divided by two
would help us find a location of the center of this ellipse.
While four is the same as 12/3, so this is 16/3,
divided by two, which is same as times 1/2,
to simplify this nicely.
So the center of this ellipse is 8/3 units
from either end of this major axis.
Well, this is 4/3, so the center would be somewhere in here,
8/3 units from the two endpoints of the major axis.
8/3 would also be the value of A.
And remember that distance from the center to the focus
is equal to C,
and since this is 8/3 and this is 4/3,
C would also equal 4/3.
And the reason this is important is remember on ellipse,
A squared is equal to B squared plus C squared.
So we can use this information to determine
the endpoints of the minor axis.
Let's go ahead and do that on the next screen.
Since we have an ellipse,
we know that A squared is equal to B squared plus C squared.
So we have 8/3 squared
equals B squared plus 4/3 squared.
So we have 64/9 wants to equal B squared plus 16/9.
Go ahead and solve for B squared.
Take the square root of both sides.
Let's go ahead and get a decimal approximation.
B is approximately equal to 2.3.
And the reason that's so important is
starting at the center,
we can go left and right approximately 2.3 units
to find the endpoints of the minor axis.
That'll help us form a nice ellipse.
And for travelling to the right, 2.3 units,
let's go out here about 2.3 units,
and then over here,
and then the same thing on the other side,
somewhere in here, out 2.3 units.
And now we can form a nice ellipse
that look something like this.
If we wanted to find the other focus,
remember the distance from here was 4/3 units.
So we could go out 4/3 units from this vertex here
to somewhere in here.
So we have the first focus and the second focus here.
Here's a computer generated graph
of the one that we just made.
And just to review,
here are the key components that were laid,
A, B, and C,
when we're graphing an ellipse.
And we also used the fact that A squared
is always equal to B squared plus C squared
when we have an ellipse.
Okay, I hope you found this video helpful.
Next, we'll take a look at graphing a hyperbola.