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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.

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  • piceratops sapling style avatar for user Ada
    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)
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    • leaf green style avatar for user Ari
      (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)
  • blobby green style avatar for user Yjj7357
    In the formula, how can it equal 1 when it has x and y value of 0?
    (36 votes)
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    • mr pants teal style avatar for user Dorus Dijkstra
      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!
      (49 votes)
  • hopper happy style avatar for user Joss Glenn
    Why do ellipses and hyperbolas equal one?
    (16 votes)
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    • leaf grey style avatar for user Nabla ∇
      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)
  • blobby green style avatar for user trepidwhlr
    @; ((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)
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    • old spice man green style avatar for user Dan Surerus
      (-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)
  • piceratops ultimate style avatar for user Mosaddek Islam
    how can we recognize what conic section it is when the equation is not written in standard form?
    (9 votes)
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    • leaf red style avatar for user Noel Kim
      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
      (8 votes)
  • leaf green style avatar for user khan.student
    If 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)
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    • old spice man green style avatar for user Dan Surerus
      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)
  • piceratops sapling style avatar for user Enrique Salazar Rios
    What is the difference between an oval and an ellipse?
    (7 votes)
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    • piceratops ultimate style avatar for user Just Keith
      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)
  • aqualine tree style avatar for user CodeLoader
    How would we make an equation of a tilted ellipse?
    (9 votes)
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  • blobby green style avatar for user arktoi
    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)
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    • old spice man green style avatar for user Dan Surerus
      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)
  • blobby green style avatar for user Hafsa Mahmood
    Why did he subtract 5 when moving in the positive direction and add 2 while moving in the negative direction?
    (3 votes)
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    • leaf green style avatar for user kubleeka
      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.