Given the conic section
using this polar equation,
we want to find the x and y intercepts and the foci.
We'll begin by determining
what type of conic section we have.
If we take the given polar equation,
and write it in this form here,
it's our equation contains cosine theta,
we'll have a conic section where e is greater than zero
is the eccentricity with a focus at the pole.
So if we write the equation in this form here
where we have a one here in the denominator,
e will give us the eccentricity,
which will help us determine which conic section we have.
So going back to our example,
we want this two to be a one,
so we'll divide the numerator and denominator by two.
We want the equation to be in the form of r equals e times d
divided by the quantity one plus e cosine theta.
So if we divide the numerator by two,
we would have 12 divided by two or six.
If we divide by the denominator by two,
two divided by two is one.
And six cosine theta divided by two is equal to three.
So we have a denominator of one plus three cosine theta.
Now that we have a one here,
we know the eccentricity is equal to three
which is greater than one and therefore we have a hyperbola.
Next, because our equation contains cosine theta
rather than sine theta, we'll have two branches
where one opens left and one opens right.
If our equation contains sine theta,
one branch would open up and one branch would open down.
Because of the equation we know
that one focus is at the pole,
which would be this point here
with coordinates zero comma zero.
We'll find the other focus later.
Now let's find the x and y intercepts
using a table of values.
We're to find the x intercepts.
We'll let theta equal zero radians and pi radians.
Since zero radians terminates along the positive x-axis,
and pi radians terminates along the negative x-axis.
And then to find the y intercepts,
we'll let theta equal pi over two radians
and three pi over two radians.
Notice pi over two radians terminates
along the positive y-axis.
And three pi over two radians terminates
along the negative y-axis.
So completing this table will give us
both the x and y intercepts.
Let's go ahead and use this form of our equation.
Notice when theta is zero radians,
cosine zero would be one,
and therefore r is equal to six divided by the quantity
one plus three times one which would be 6/4 or 3/2.
So when theta is zero radians, r is equal to 3/2,
and again, zero radians terminates here
along the positive x-axis.
So if r is 3/2 or 1.5, we'd have a point here on the x-axis
where the coordinates would be 3/2 comma zero
using Cartesian coordinates.
Next, when theta is pi radians,
notice cosine pi would be equal to negative one,
so we'd have r equals six divided by
the quantity one plus three times negative one,
this would be six divided by negative two or negative three.
So when theta is pi radians, r is equal to negative three.
Pi over two radians terminates
along the negative x-axis here,
but since r is negative three,
we plot the point in the opposite direction
which would be this point here.
So the other x intercept is the point three comma zero.
And now to find the y intercepts.
When theta is pi over two, cosine pi over two is zero,
so r would just be six divided by
the quantity one plus three times zero
which would be just six.
So when theta is pi over two, r is six.
Pi over two radians terminates here
along the positive y-axis.
And since r is six, this would be the point on the curve
which is a y intercept with coordinates zero, six.
And then finally when theta is three pi over two radians,
cosine three pi over two radians is also zero.
So we'd have the same r value of six.
So when theta is three pi over two radians,
it terminates here along the negative y axis,
r is positive six which would give us this point here,
zero comma negative six.
Which means one branch of the hyperbola passes
through these three points,
and the other branch passes through this point
and looks something like this.
We're not quite done yet.
We only have one focus.
There's another focus over here on the right.
To help us find the other focus,
let's begin by determining the center
which would be the point half way
between the two vertices somewhere in here.
To find the coordinates of the center,
we'd have to find the average of the x coordinates
of these two x intercepts.
So we'll add these x coordinates and divide by two
or add them and multiply by 1/2.
So 1/2 times 3/2 plus three or three over one,
we'd have a common denominator of two.
So we have 1/2 times 3/2 plus 6/2, that's 9/2.
So 9/4 would be the x coordinate of the center.
So we have 9/4 comma zero,
since the point does lie on the x-axis.
And the reason we need this point is
if we go back to our notes,
the distance from the center to a focus
is equal to c units.
So if we find the distance from the center to one focus,
then we can find the coordinates of the second focus
over here on the right.
Again, this distance here,
we'll call little c, would be the same distance
to the other focus here on the right.
And since this first focus we'll call F sub one
has coordinates zero comma zero,
because the y coordinates are zero,
notice the distance between these two points would be
9/4 minus zero or just 9/4 units.
Which means to find the focus on the right,
we would add 9/4 to the x coordinate of the center.
So the second focus, F sub two would have an x coordindate
of 9/4 plus c or plus 9/4 comma zero.
F is equal to 18/4 or 9/2 comma zero.
So the other focus would have coordinates, again,
9/2 comma zero.
We know that because if we add c units
to the x coordinate of the center,
it brings us out to 9/2 comma zero or 4.5 comma zero
which is here.
And that's because this distance from c to F
has to be c units.
I hope you found this helpful.