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## Algebra (all content)

### Course: Algebra (all content) > Unit 17

Lesson 11: Identifying conic sections from their expanded equations# Conic section from expanded equation: ellipse

Sal manipulates the equation 9x^2+4y^2+54x-8y+49=0 in order to find that it represents an ellipse. Created by Sal Khan.

## Want to join the conversation?

- what about in the last equation - it equals 1 but if you have x and y -3 and 1 doesn't it have to be 0?(21 votes)
- The point (-3, 1) is not on the graph. It is a key point for the graph (the center) but you will notice that the actual graph (the perimeter of the circle) does not cross the point (-3,1).

This is similar to having the graph of y=x and saying that if you put in 10 for x and 1 for y you end up with 1=10, which isn't true but that's fine because the point (1,10) isn't on the graph of y=x.(29 votes)

- Where can i find the videos about equations of elipses, circles and parabolas? I'm afraid i don't really understand those completely.(11 votes)
- see introduction to conic sections to get started...(1 vote)

- Was this example just a coincidence? The coefficent on the y-term ended up under the x-term and the coefficent on the x-term ended up under the y-term.(10 votes)
- Yes, it was just a coincidence, as the two happened to multiply to 36, the number that both were divided by.(7 votes)

- Is it just me, or do you ever stop after finishing an equation as long and complicated as this and just feel really intelligent? I know in the long run this equation isn't super hard, but the feeling of completing something that crazy-looking is super empowering.(4 votes)
- I understand that an equation for an ellipse is always set equal to one, not zero. Why is this?(4 votes)
- You could subtract 1 from both sides of the equation, and then it would be set equal to zero.

It is just one of several conventions for the equations of circles, ellipses, and hyperbolae to be presented in this form, whereas the equations of parabolae tend to be presented in the form ax² + bx + c = 0.

However, the general form for the equation of any conic section is:

Ax² + By² + Cxy + Dx + Ey + F = 0

Therefore, depending on context, you may see different conventions followed.

For an ellipse, there does need to be a non-zero constant term in order to obtain a curve of any kind (as opposed to a single point). Whether you place the constant on the left or right side of the equation is a matter of taste, but the x² and y² terms are both necessarily positive, so a constant on the oposite side of the equation must be positive, while if placed on the same side it would have to be negative, otherwise there would be no real solutions to the equation, and therefore no points on the coordinate plane could be plotted and there would be no ellipse.(1 vote)

- How do I know if an equation is representing an ellipse but not a circle?(2 votes)
- An easy way to tell the difference between an ellipse and a circle is if their radii are the same when the equation is in standard form (the way it was after Sal completed the square). For example, if the number under the (x-h)^2 and the number under the (y-k)^2 are equal, then you have a circle. If they are not equal, the radii are different lengths, so the equation is an ellipse.(5 votes)

- How exactly, though, does Sal know that it is an ellipse? By just looking at it? I think I need a bit of clarification in terms of knowing what figures are which by just looking at the equation. Thanks.(2 votes)
- In the conic sections -

Hyperbola - the 2nd degree variables(x^2 and y^2) need to have opposite signs

Parabola - needs to have only 1 of the variables(x or y) as square while the other is degree 1(just x or y).

Ellipse - needs to have both variables in degree 2

Circle - special ellipse

Looking at the above terms we can easily rule out Parabola and Hyperbola. So Sal says that it is probably an ellipse.(3 votes)

- At the beginning of the video on Identifying Conics 1, how did he know to complete the square in order to get it into standard form? Does that just come from experience or is there a way to tell?(2 votes)
- if you are given an equaiton like the one in the video, the first action to be taken is completing it to the square. To complete the squre will make you to see what you are dealing with easly.(3 votes)

- what about in the last equation - it equals 1 but if you have x and y -3 and 1 doesn't it have to be 0?(1 vote)
- That's the center. It equals 1 when it is on the conic section but the center is never on the shape(3 votes)

- at4:10, where did you get the 81 to add to the right hand side?(1 vote)

## Video transcript

The standard question you often
get in your algebra class is they will give you this
equation and it'll say identify the conic section and
graph it if you can. And the equation they give you
won't be in the standard form, because if it was you could
just kind of pattern match with what I showed in some of the
previous videos and you'd be able to get it. So let's do a question
like and let's see if we can figure it out. So what I have here is 9x
squared plus 4y squared plus 54x minus 8y plus
49 is equal to 0. And once again, I mean who
knows what this is it's just not in the standard form. And actually one quick clue to
tell you what this is you look at the x squared and the y
squared terms if there are. If there's only an x squared
term and then there's just a y and not a y squared term, then
you're probably dealing with a parabola, and we'll go
into that more later. Or if it's the other way
around, if it's just an x term and a y squared term,
it's probably a parabola. But assuming that we're dealing
with a circle, an ellipse, or a hyperbola, there will be an x
squared term and a y squared term. If they both kind of have the
same number in front of them, that's a pretty good clue
that we're going to be dealing with a circle. If they both have different
numbers, but they're both positive in front of them,
that's a pretty good clue we're probably going to be
dealing with an ellipse. If one of them has a negative
number in front of them and the other one has a positive
number, that tells you that we're probably going to be
dealing with a hyperbola. But with that said, I mean that
might help you identify things very quickly at this level, but
it doesn't help you graph it or get into the standard form. So let's get it in
the standard form. And the key to getting it in
the standard form is really just completing the square. And I encourage you to re-watch
the completing the square video, because that's all we're
going to do right here to get it into the standard form. So the first thing I like to do
to complete the square, and you're going to have to do it
for the x variables and for the y terms, is group
the x and y terms. Let's see. The x terms are 9x
squared plus 54x. And let's do the y
terms in magenta. So then you have plus 4y
squared minus 8y and then you have-- let me do this in a
different color-- plus 49 is equal to 0. And so the easy thing to do
when you complete the square, the thing I like to do is, it's
very clear we can factor out a 9 out of both of these numbers,
and we can factor out a 4 out of both of those. Let's do that, because
that will help us complete the square. So this is the same thing
is 9 times x squared plus 9 times 6 is 54, 6x. I'm going to add something
else here, but I'll leave it blank for now. Plus 4 times y squared minus
2y I'm probably going to add something here too, so I'll
leave it blank for now. Plus 49 is equal to 0. So what are we
going to add here? We're going to
complete the square. We want to add some number here
so that this whole three term expression becomes
a perfect square. Likewise, we're going to add
some number here, so this three term number expression
becomes a perfect square. And of course whatever we add
on the side, we're going to have to multiply it by
9, because we're really adding nine times that. And add it on to that side. Whatever we add here, we're
going to have to multiply it times 4 and add
it on that side. If I put a 1 here, it's really
like as if I had a 4 here, because 1 times 4 is 4 and if I
had a 1 here it's 1 times 9. So 9 there. Let's do that. When we complete the square,
we just take half of this coefficient. This coefficient is 6, we
take half of it is 3, we square it, we get a 9. Remember it's an equation, so
what you do to one side, you have to do to the other. So if we added a 9 here, we're
actually adding 9 times 9 to the left-hand side of the
equation, so we have to add 81 to the right-hand side to make
the equation still hold. And you could kind of view
it if we go back up here. This is the same thing, just
to make that clear as if I added plus 81 right here. Of course I would have had
to add plus 81 up here. Now let's go to the y terms. You take half of this
coefficient is minus 2, half of that is minus 1. You square it, you get plus 1. 1 times 4, so we're really
adding 4 to the left-hand side of the equation. And just so you understand
what I did here. This is equivalent as if I just
added a 4 here, and then I later just factored out this 4. And so what does this become? This expression
is 9 times what? This is the square of-- you
could factor this, but we did it on purpose-- it's x plus 3
squared and then we have plus 4 times-- What is
this right here? That's y minus 1 squared. You might want to review
factoring of polynomial or completing the square if
you found that step a little daunting. And then we have plus 49 is
equal to 0 plus 81 plus 84 is equal to 85. All right, so now we have 9
times plus 3 squared plus 4 times y minus 1 squared. And let's subtract
49 from both sides. That is equal to-- let's see if
I subtract 50 from 85 I get 35, so if I subtract 49, I get 36. And now we are getting close to
the standard form of something, but remember all the standard
forms we did except for the circle-- we had a y-- and we
know this isn't a circle, because we have these weird
coefficients, well not weird but different coefficients
in front of these terms. So to get the 1 on the
right-hand side let's divide everything by 36. If you divide everything by 36,
this term becomes x plus 3 squared over see 9 over 36 is
the same thing as 1 over 4, and then you have plus y minus 1
squared 4 over 36 is the same thing as 1 over 9 and all
of that is equal to 1. And there you go. We have it in the standard
form, and you can see our intuition at the beginning
the problem was correct. This is indeed an ellipse, and
now we can actually graph it. So first of all, actually
good place to start, where's the center of the this
ellipse going to be? It's going to be x is
equal to negative 3. What x value makes
this whole terms 0? So it's going to be x is
equal to minus 3, and y is going to be equal to 1. What y value makes this
term 0? y is equal to 1. That's our center. So let's graph that, and then
we can draw the ellipse. It's going to be in the
negative quadrant. This is our x-axis and
this is our y-axis. And then the center of our
ellipse is at minus 3 and positive 1, so
that's the center. And then, what is the
radius in the x direction? We just take the square
root of this, so it's 2. So in the x direction we
go two to the right. We go two to the left. And in the y direction,
what do we do? Well we go up three
and down three. The square root of this. Let me do that. Remember you have to take the
square root of both of those. The vertical axis is actually
the major radius or the semi-major axis is 3, because
that's the longer one. And then the 2 is the minor
radius, because that's the shorter one. And now we're ready to
draw this ellipse. I'll draw it in brown. Let me see if I can
do this properly. I have a shaky hand. All right, it looks
something like that. And there you go. We took this kind of crazy
looking thing, and all we did is algebraically manipulate it. We just completed the squares
with the x's and the y terms. And then we divided both sides
by this number right here and we got it into the
standard form. We said oh this is an ellipse. We have both of these terms,
they're both positive, we're adding we're not subtracting,
they have different coefficients underneath here. So we're ready to go over the
ellipse, and we realized that the center was at minus 3,1,
and then we just drew the major radius, or the major
axis and the minor axis. See you in the next video.