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# Intro to ellipses

Sal introduces ellipses and shows how their standard equation relates to their center and radii (ellipses have two radii: major and minor). Created by Sal Khan and NASA.

## Want to join the conversation?

- What if I have an equation that has coefficients? For example:

(25x^2) + (16x^2) = 400

Would a^2 then be equal to 1/25? And what would I do with the 400?(52 votes)- (I'm assuming your second x is supposed to be a y, otherwise this is just an algebraic equation)

you first divide both sides by 400 to get (25x^2)/400 + (16y^2)/400 = 1

then rearrange that to get (x^2/400/25) + (y^2/400/16) = 1

a^2 = 400/25 = 16

a=4

b^2 = 400/16 = 25

b=5(89 votes)

- In the formula, how can it equal 1 when it has x and y value of 0?(36 votes)
- I was also struggling with that question, but after some thinking it became pretty clear. The ellipse is constructed out of tiny points of combinations of x's and y's. The equation always has to equall 1, which means that if one of these two variables is a 0, the other should be the same length as the radius, thus making the equation complete. Which is exactly what we see in the ellipses in the video.

We find the center by making x and y equal 0, but not at the same time:

This is what happens:

When x=0 then y=1 (and remember that r=1 always!)

x^2+y^2=r^2, so 0 + 1 = 1

When y=0 then x=1

x^2+y^2=r^2, so 1 + 0 = 1

The equation itself doesn't match (0,0) (only if r=0, which is never the case), but the above method gives us a way to search the exact center of the ellipse. It's quite as simple as that. I hope my explanation helps you (and maybe others) to understand the concept behind this equation!(50 votes)

- Why do ellipses and hyperbolas equal one?(16 votes)
- Because they are 'deformations' of a circle. You know what an equation of a circle looks like, right? Suppose you have this one:

(x - 3)² + (y - 2)² = 4

The center is at (3, 2), but how can one squeeze the circle to make it appear as an ellipse or hyperbola? Divide both side by 4 and you get:

(x - 3)²/4 + (y - 2)²/4 = 1

It's still the same circle, but now you know where that 1 is coming from and you can also squeeze it by changing the denominator of either the x or y term. In this video you can see what happens when the denominator of one of the terms changes:

https://www.screenr.com/lBZN

The equation 'd' is the one I've written above and equation 'e' is:

(x - 3)²/4 + (y - 2)²/**b**= 1

Where**b**is the variable that we're changing. Notice that when b = 4, it forms the same circle as 'd', but when b =/ 4 and still positive it's an ellipse. When it goes to negative, it becomes a hyperbola.(20 votes)

- @13:00; ((y-1^2)/4)+((x+2^2)/9)=does not eqUAL 1? Am I suppose to know right off the bat that the problem has to do with ellipses (shifting). First of all, when I substituted in (-2,1) for the x and y into the equation, it didn't equal 1. I tried what you're suppose to follow for order of operations, and that gave me 0/4+0/9, which is not equal to zero, I even tried factoring just to see if it would work, and it didn't: 0/4+8/9.

(13 votes)- (-2,1) is the center of the elipse, it will not satisfy the equation because it is not ON the elipse. Ive typed the reasons in previous questions, read all those and you should get it.

Put in the center coord. for y and solve for x to get your 2 x direction points, (x^2+4x+4=9 ... x=1 and x=-5 so (1,1) and (-5,1) are 2 major points on the elipse. Do the same for your X center coord. and solve for Y. OR once you get used to the standard form of the ellipse you can see that a^2=2 and b^2=3, so you can just use that to add and subtract 2 from your y coord of center point and add and subtract 3 from your x coord of center point to get the same information.

Yes, after awhile you will notice the general equations for conic sections with practice. You will get some equation on a test that doesn't look like a conic at first, but you need to do algebraic manipulations like completing squares to put the given information into the standard forms of ellipses, parabolas, hyperbolas...

Hope that helps...(16 votes)

- how can we recognize what conic section it is when the equation is not written in standard form?(9 votes)
- For a parabola: there is only one squared term in the equation

Circle: the coefficents are the same for the two squared terms

Ellipses: the two fractions are added together

Hyperbola: the two fractions are subtracted from each other(7 votes)

- What is the difference between an oval and an ellipse?(7 votes)
- The term "oval" isn't really used much in geometry because it does not have a very clear definition. Thus, you'd have to ask for clarification if someone mentioned an oval in geometry.

Since oval does not have a standard exact meaning in mathematics, we cannot really compare it to an ellipse which does have a clear meaning.(7 votes)

- 9:40If you were to shift the ellipse down 2 spaces, why would you add 2 to y? Wouldn't you subtract 2 from y? Why is it like the opposite of what I think? In other words, why would you want the numerator to equal zero? This seems counter-intuitive.(10 votes)
- Think about where your center point is. In this example it is C(5,-2). If you make the numerator zero, by putting in your Y coord. of your center point, it cancels the Y term out of your equation so that you can solve for your x values.

(x-5)^2=9

x^2-10x+25=9

x^2-10x+16=0

(x-2)(x-8)=0 .... so x=8 or x=2

You can use your center point that you put in for y which was -2 and these x's

(8,-2) and (2,-2)

These are now your 2 points on the ellipse on either side of the center point in the X directions.

If you put in (y-2)^2/25 instead: Try to put in your Y center coord now and see what you end up with. (-2-2)^2/25 = (16/25) so your Y term doesnt cancel out and allow you to find the x direction major axis points on the ellipse.

Think about just a circle x^2+y^2 = 1

To find your 2 x intercepts you want Y to equal zero. Put in Y=0 and you have x^2=1 .. so x=1 or x=-1 and your points on your x axis are

(1,0) and (0,1)

Same thing if you shift the Y coord of the center down 2:

x^2+(y+2)^2=1

Here your center is (0,-2) from shifting down 2, and notice you put a (y+2) term in because you want the Y term to cancel by becoming zero when you put the center point's Y term into the equation.

Hope that helps, i can clarify more if you still dont get it

Pantera62345@hotmail.com(5 votes)

- How would we make an equation of a tilted ellipse?(9 votes)
- other than the original 4 points, i don't see how to figure out any other point on the shifted ellipse. can somebody just show me how to find one other point somewhere on it?(4 votes)
- If you know the equation of the ellipse, pick an x value between -a and a and figure out what y is. OR pick a y value between -b and b and find out x. in the problem on the vid (the blue one), a=2 and b=3 so pick an x between -2 and 2 and plug it into the equation and figure out Y. there you have a point (x,y) on the ellipse.(4 votes)

- Why did he subtract 5 when moving in the positive direction and add 2 while moving in the negative direction?(3 votes)
- In the original equation, letting x=0 gives us a certain point. We want to shift the figure to the right by 5. That is, we want x=5 in the new equation to give the same point as x=0 in the original. So we replace x with x-5, because plugging in 5 to this equation gives 0.(4 votes)

## Video transcript

In the last video, we learned a
little bit about the circle. And the circle is really just
a special case of an ellipse. It's a special case because in
a circle you're always an equal distance away from the center
of the circle, while in an ellipse, the distance from the
center of the circle is always changing. You know what an
ellipse looks like. Well, I showed you that
in the first video. It looks something like that. What I mean is that the radius
or the distance from the center is always changing. Let me say this is
centered at the origin. So that's the origin
right there. You see here, we're really, if
we're on this point on the ellipse, we're really
close to the origin. This is actually the closest
we'll ever get, just as close as well get down here. And when we're out here we're
really far away from the origin and that's about as far as
we're going to get right there. So a circle is a special case
of this, because in a circle's case, the furthest we get from
the origin is the same distance as the closest we get, or, in
other words, we are always the exact same distance
away from the origin. Well, with that said, let's
actually go a little bit into the math. So the general or the standard
form for an ellipse centered at the origin is x squared over a
squared plus y squared over b squared is equal to 1. Where a and b are just
any two numbers. I could have written this as
c squared and d squared. I mean, they're just
place holders. Just to give you an idea of
what this means, if this was our ellipse in question right
now, a is the length of the radius in the x-direction. Remember, we're going to
have a squared down here. So if you took the square
root of whatever is in the denominator, a is the x-radius. So this distance in our little
chart right here, in our little graph here, that distance is a,
or that this point right here, since we're centered at the
origin, will be the point x is equal to a y is equal to 0. And of course this point right
here this will be a, so this would be the point
minus a comma 0. And then the radius in the
y-direction would be this radius right here and is b. So this point would be x is
equal to 0, y is equal to b. Likewise this point right here
would be x is equal to 0, y is equal to minus b. And the way I drew this, we
have kind of a short and fat ellipse you can also have kind
of a tall and skinny ellipse. But in the short and fat
ellipse, the direction that you're short in that's
called your minor axis. And so b, I always forget the
exact terminology, but b you can call it your semi or the
length of your semi-minor axis. And where did that
word come from? Well if this whole thing is
your minor axis or maybe you could call your minor diameter
if this whole thing is your minor diameter it's called
minor, because it's the shortest of all of the
diameters of this ellipse. And then the semi
is half of that. b is the length of
the semi-minor axis. That's b in this example, just
because as I drew this ellipse it just happens to be that
b is smaller then a. If b was larger than a, I would
have a tall and skinny ellipse. Let me actually draw one. It could have been like this. I could have an ellipse that
looks something like that. In which case, all of a sudden
b would be the semi-major axis, because b would be
greater than a. That this would be
taller than it is wide. But let me not confuse
the graph too much. And in this case, a is the
length of-- I think you've guessed it-- a is the length of
the semi-major axis or you can even call it the length
of the major radius. I think that makes more sense. And you can call this
the minor radius. So let's just do an example. And I think when I've done an
example with actual numbers, it'll make it all a
little bit clearer. So let's say I were to show up
at your door with the following: If I were to say x
squared over 9 plus y squared over 25 is equal to 1. So what is your radius
in the x-direction? This is your radius in
the x-direction squared. So your radius in the
x-direction if we just map it, we would say that
a is equal to 3. Because this is a squared. And if we were just map it we'd
say this is b squared than this tells us that b is equal to 5. So if we wanted to graph
this, and once again this is centered at the origin. Let me draw the ellipse first. So, first of all, we have our
radius in the y-direction is larger than our radius
in the x-direction. The ellipse is going to
be taller and skinnier. It's going to look
something like that. Draw some axes, so that could
be your x-axis, your y-axis. This is your radius
in the y-direction. So this distance right here
is going to be 5, and so will this distance. And this is your radius
in the x-direction. So this will be 3,
and this will be 3. That's it. You have now plotted
this ellipse. Nothing too fancy about it. And actually just to kind of
hit the point home that the circle is a special
case of an ellipse. We learned in the last video
that the equation of a circle is x squared and a circle
centered at the origin. x squared plus y squared
is equal to r squared. So if we were to divide both
sides of this by r squared, we would get-- and this is just
little algebraic manipulation-- x squared over r squared plus y
squared over r squared is equal to 1. Now in this case, your a
is r and so is your b. So your semi-minor axis
is r and so is your semi-major axis of r. Or, in other words, this
distance is the same as that distance, and so it will
neither be short and fat nor tall and skinny. It'll be perfectly round. And so that's why the circle is
a special case of an ellipse. So let me give you a slightly--
It'll look a lot more complicated, and this is
something you might see on exam. But I just want to show you
that this is just a shifting. Let's say we wanted to
shift this ellipse. Let's say we wanted to shift
it to the right by 5. So instead of the origin being
at x is equal to 0, the origin will now be at x is equal 5. So a way to think about that is
what does this term have to be so that at 5 this term
ends up being 0. Well I'll actually draw it
for you, because I think that might be confusing. So if we shift that over the
right by 5, the new equation of this ellipse will be x minus 5
squared over 9 plus y squared over 25 is equal to 1. So if I were to just draw
this ellipse right now, it would look like this. I want to make it look
fairly similar to the ellipse I had before. It would look just like that. Just shifted it over by five. And the intuition we learned a
little bit in the circle video where I said, oh well, you
know, if you have x minus something that means that the
new origin is now at positive 5. And you could memorize that. You could always say, oh, if I
have a minus here, that the origin is at the negative of
whatever this number is, so it would be a positive five. You know, if you had a positive
it would be the opposite that. But the way to really think
about it is now if you go to x is equal to 5, when x is equal
to 5, this whole term, x minus 5, will behave just like
this x term will here. When x is equal to 5 this
term is 0, just like when x was 0 here. So when x is equal to 5, this
term is 0, and then y squared over 25 is equal 1, so y
has to be equal five. Just like over here when x is
equal is 0, y squared over 25 had to be equal to 1,
y is equal to either positive or minus 5. And I really want to give
you that intuition. And then, let's say we
wanted to shift this equation down by two. So our new ellipse looks
something like this. A lot of times you learned
this in conic sections. But this is true any function. When you shift things,
you shift it this way. If you shift this graph to the
right by five, you replace all of the x's with x minus 5. And if you were to shift it
down by two, you would replace all the y's with y plus 2. So let me draw our new
ellipse first, just to show you what I'm doing. So our new ellipse is going
to look something like that. I'm shifting the yellow
ellipse down by two. So this equation, if I shift it
down, well, the x is still where it was before. x minus 5
squared over 9 plus y plus 2 squared over 25 is equal to 1. And once again, the reason I
know this is because now when y is minus 2, this
whole term is 0. 0 when y equals minus 2. And when this term is 0, it
behaves the same way as when this term was 0. So when y is equal to minus 2,
you get the same behavior, you're at the same point in the
curve, right here actually, as you are when y equaled 0
in this one, so here. So it's not the same point. You can kind of view it as the
same part of the ellipse. You're at kind of the maximum
width point on the ellipse here and here when y is equal to 2,
and you were here at y equal to 0-- sorry, when y
equals minus 2. This is minus 2. And that's because when you
put y equals minus 2 here this whole term is 0. Just like when y was 0 here. I don't want to make
it too confusing. But just to kind of wrap it all
up, sometimes you might see something like graph the
following: y minus 1 squared over 4 plus x plus 2 squared
over 9 is equal to 1. And so the first thing you
could say is OK this is just like the standard ellipse y
squared over 4 plus x squared over 9 is equal to one. It's just like this,
but it's shifted over. This ones origin is 0,0, while
this ones origin would be what? It would be the point x
is minus 2 and y is 1. So if you were to graph
this, your radius in your y-direction is 2. 2 squared is equal to 4. Your radius in your
x-direction is 3. 3 squared is equal to 9. So your x-radius is actually
larger than your y-radius. So, it's going to be a little
bit of a fat ellipse. Actually, let me draw
the axes first. Let me draw it like this. That's my vertical axis,
this is my x-axis. And so my center is
now at minus 2, 1. That's minus 2,
and I go up one. That's the center
of my ellipse. And now in the x-direction,
this is the x term, my x-radius is 3. So the ellipse will go
three-- in that direction. This is it's widest point
will be 3 in that direction. And then in the y-direction,
it'll go 2, so it'll go up 1, 2 so that's there and
then 1, 2 and it's there. So if I were to draw that
ellipse it would look something like this through my best shot. A little bit fatter than it
is tall, and that's because your x-radius is larger
than your y-radius. This distance right here is 3,
this distance right here is 3, this distance right here is 2,
this distance right here is 2. You could figure out
what these points are. I won't do all of them right
now just for the sake of time. But this right here is
the point minus 2, 1. So if you go three more than
that-- so if you add 3 to the x-direction this is
the point 1 comma 1. If you would take three away
from the x-direction, this would be minus 5 comma 1. And you could figure
out the other points. That might be good
exercise for you. Anyway that's a little
bit on ellipses. In future videos we'll do
really hairy problems where you have to simplify it into this
form so that we know that it definitely is an ellipse.