Now
that you know what the conic sections are and why they're
called conic sections, let's see if we can understand the
actual equations of the conic sections a little bit better,
and use that knowledge to be able to at least recognize them
when we see the equation, and then if we see the equation,
know how to plot them.
So the first one we'll start off with is the circle.
And you've probably seen this for many years already,
but I'll review, if you don't know it already.
The general form of an equation, or the general
equation for a circle, is x squared plus y squared is equal
to r squared, where r is the radius of the circle.
And this would be a circle centered at the point 0,0.
So I'll just draw this kind of a general graph right here.
So that's the x-axis, that's the y-axis, the circle will
look like this-- let me do it in a different color-- the
circle would look something like-- make sure it's a
circle-- well, it's supposed to be centered at 0,0, but that's
close enough-- so its center will be right here, and the
radius if you go from the center to any point along that
circle will have a distance of r, so if you go from there to
there it's r, from there to there it's r, from there to
there it's r, and to some degree, this formula, all it is
is an extension of the distance formula, which is really just
the extension of the Pythagorean Theorem.
So for example, the distance formula, if I want to know the
distance between some point x,y and the point 0,0, what you do
is, you take the difference of the x's-- so x minus 0-- you
square that, and then you add that to the distance between
the y's squared-- so that's one y point minus 0y-- y is equal
to 0-- square that, and that is equal to the distance squared.
So if you simplify this, x minus 0 squared, that's just
x squared plus-- and this is just y squared is equal
to distance squared.
So essentially, this equation is the plot of all points that
are exactly d away, a distance of d away, from the point 0,0,
and that's just a circle.
And I'll let you think about it, I think I actually showed
this to you in a distance formula video, but the distance
formula just comes out of the Pythagorean Theorem.
And if that's not completely obvious for you, just think
about a little bit, and it'll hopefully become a
little bit more obvious.
But anyway, this was probably-- you probably already knew this,
and actually just to hit the point home, if we had the
equation like this x squared plus y squared is equal to 9,
the graph of this circle would look like this.
So that's the x-axis, that's the y-axis, then draw the
circle itself, circle looks like-- close enough, and then
the distance or the radius from the center of the circle,
that's going to be 3.
There's a 9 here, why isn't the radius 9?
Oh, that's because this is the radius squared.
Remember the original formula I just showed you. x squared plus
y squared is equal to r squared.
So this right here is r squared, so if r squared
is equal to 9, than r is equal to 3.
It can't be minus 3.
I mean, it could, you can't have a negative radius, or if
you did you're just going in the other direction, but
it's the same thing.
So the radius is equal to 3.
So that's a circle, and that's pretty straightforward, but in
a lot of algebra classes, they complicate the issue a little
bit by shifting the circle.
So let's just shift this circle.
Instead of having-- so let me just rewrite it.
So the unshifted circle was x squared plus y squared is equal
to-- let me write it this way-- is equal to 3 squared, that's
the same thing as 9, and let's say that the new circle, the
shifted circle, is x minus 1 squared plus y plus 2 squared
is equal to 3 squared.
Now all of a sudden this looks really complicated and daunting
and all the rest, but all you have to recognize is, we just
substituted an x minus 1 for the-- whoops, messed
up my pointer.
We just substituted an x minus 1 for the x, and
we just substituted a y plus 2 for the y.
So it has the same basic pattern as of this circle, and
the fact that we added or subtracted numbers from the x's
and the y's tells us that we shifted the circle, and so the
next obvious question is, where did you shift it to?
And your impulse might be, oh, well maybe I shifted it to,
instead of the center being 0,0, your intuition might be
to say that, well, now the center is at negative 1,2.
And you'd be almost right, except you'd be exactly the
opposite of the correct answer.
The new center is now x is equal to positive 1 and
y is equal to minus 2.
And that might be unintuitive for you at first-- and you
might want to watch some of the videos, I think I've done them
already, or I've always intended to, on shifting
functions-- but the way to think about it is the center
here is x is equal to 0.
So when x and y is equal to 0, x squared plus y squared is 0,
you're exactly 0 away from the center, or we're at the center.
So now if we want to be-- if we want x to be 0 away from our
new center, this term has to be equal to 0.
And if-- just like when x was equal to 0, this term equaled
0, so now for us to be at the center of our new circle, so to
speak, this term has to be 0.
So the new center has to be at x is equal to 1.
Likewise, this has to be 0, and so the center
is at y is equal to 2.
Another way to think about it is this-- let's say when y,
you know what happens here when y is equal to 2.
Whatever part of the circle we're in when y is equal to 2.
There's some part of the circle we're at, in fact I could draw
it, when y is equal to 2.
Let's say this radius is 3 when y is equal to 2, we're probably
right around there on the circle.
We could be there or we could be there.
Now we're shifting the circle, so now instead of being at 0,0,
we're going to be at 1 minus 2.
So now we're going to be at the new center is x is equal to 1,
y is equal to minus 2, the new center is there, and if I were
to draw the new circle, it would look something like this.
I'm going to try my best to draw it still as a
circle and show you that it's been shifted.
No, that's not good.
Let me draw it like-- I pressed the wrong button.
That's not good.
Let me draw it right there.
That's close enough.
I don't have to keep doing it.
So what we did is we shifted this circle down 2,
and to the right 1.
So if we take its center point, we went down 2
and to the right 1.
And so if you think about up here, when y was equal to 2, we
could have been at this point or this point, the kind of
equivalent points of the new circle are going to be here.
Are going to be roughly here, where you're shifting it
down and to the right.
And in order to have that same behavior in the circle
there, this whole thing should be equal to 2.
So that same point on the circle, if this whole thing is
going to be equal to 2-- because it's going to be the
same kind of behavior in the equation, and I hope I'm not
confusing you there-- then the new y has to be 0, and
you see it there.
At both of these points now, y is equal to 0.
So I know that's a little unintuitive, but I want you to
sit and think about that a lot.
I mean, you could just memorize it, that it's the opposite,
when you have x minus 1 and y plus 2 that it's actually you
shifted to x is equal to-- the center is now 1, minus 2, or
you could memorize, if you like, that what makes this 0
and what makes this 0, and that's your new center.
But I really want you to think about it, this
is really a shift.
And of course, if you were to graph it, you were to
get this thing up there.
Anyway, let me see how much time I have.
Actually I didn't even keep track of the time.
I'll leave you there, I'll continue this in the next video
where I'll talk a little bit about ellipses.