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Now, I said all I want is a y on the left-hand side, so let's divide everything by two times b minus k. So let's divide everything times two times b minus k. So two times b minus k, and I'm actually going to divide this whole thing by two times b minus k. Now, obviously, on the left-hand side, this all cancels out. You're left with just a y, and then it's going to be y equals, y is equal to one over two times b minus k, and notice b minus k is the difference between the y-coordinate of the focus and the y-coordinate, I guess you could say, of the line, y equals k. So it's one over two times that, times x minus a squared. x minus a squared. So if you knew what b minus k was, this would just simplify to some number, some number that's being multiplied times x minus a squared. So hopefully this is starting to look like the parabolas that you remember from your childhood, if you do remember parabolas from your childhood. All right, so then let's see if we could simplify this thing on the right, and you might recognize b squared minus k squared, that's the difference of squares. That's the same thing as, that's the same thing as b plus k times b minus k. Times b, whoops, times b minus k. So the b minus k's cancel out, and we are just left with, and we deserve a little bit of a drum roll, we're just left with 1 1 2 times b plus k. So there you go. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
So if you knew what b minus k was, this would just simplify to some number, some number that's being multiplied times x minus a squared. So hopefully this is starting to look like the parabolas that you remember from your childhood, if you do remember parabolas from your childhood. All right, so then let's see if we could simplify this thing on the right, and you might recognize b squared minus k squared, that's the difference of squares. That's the same thing as, that's the same thing as b plus k times b minus k. Times b, whoops, times b minus k. So the b minus k's cancel out, and we are just left with, and we deserve a little bit of a drum roll, we're just left with 1 1 2 times b plus k. So there you go. Given a focus at the point a comma b, and a directrix at y equals k, we now know what the formula of the parabola is actually going to be. So for example, so for example, if I had a focus, if I had a focus at the point, I don't know, let's say the point one comma two, and I had a directrix, I had a directrix at y, y is equal to, I don't know, let's make it y is equal to negative one, what would the formula, or what would the equation of this parabola be? Well, it would be y is equal to one over two times b minus k. So two minus negative one, that's the same thing as two plus one, so that's two times, that's just three, two minus negative one is three, times x minus, times x minus one, whoops, x minus one squared plus 1 1 2 times b plus k. Two plus negative one is one. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
That's the same thing as, that's the same thing as b plus k times b minus k. Times b, whoops, times b minus k. So the b minus k's cancel out, and we are just left with, and we deserve a little bit of a drum roll, we're just left with 1 1 2 times b plus k. So there you go. Given a focus at the point a comma b, and a directrix at y equals k, we now know what the formula of the parabola is actually going to be. So for example, so for example, if I had a focus, if I had a focus at the point, I don't know, let's say the point one comma two, and I had a directrix, I had a directrix at y, y is equal to, I don't know, let's make it y is equal to negative one, what would the formula, or what would the equation of this parabola be? Well, it would be y is equal to one over two times b minus k. So two minus negative one, that's the same thing as two plus one, so that's two times, that's just three, two minus negative one is three, times x minus, times x minus one, whoops, x minus one squared plus 1 1 2 times b plus k. Two plus negative one is one. So one. And so what's this going to be? You're going to get y is equal to one over 1 6th x minus one squared plus 1 1 2. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
Well, it would be y is equal to one over two times b minus k. So two minus negative one, that's the same thing as two plus one, so that's two times, that's just three, two minus negative one is three, times x minus, times x minus one, whoops, x minus one squared plus 1 1 2 times b plus k. Two plus negative one is one. So one. And so what's this going to be? You're going to get y is equal to one over 1 6th x minus one squared plus 1 1 2. There you go. That is the parabola with a focus at one comma two and a directrix at y equals negative one. Fascinating. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
And it's a pretty neat little widget here, because what I can do is I can take this dot and I can move it around to redefine the center of the circle. So it's centered at 3, negative 2. So x is 3 and y is negative 2, so that's its center. It has to have a radius of 5. The way it's drawn right now, it has a radius of 1. The distance between the center and the actual circle, the points that define the circle, right now it's 1. I need to make this radius equal to 5. | Graphing circles from features Mathematics II High School Math Khan Academy.mp3 |
It has to have a radius of 5. The way it's drawn right now, it has a radius of 1. The distance between the center and the actual circle, the points that define the circle, right now it's 1. I need to make this radius equal to 5. So let's see if I take that. So now the radius is equal to 2, 3, 4, and 5. There you go. | Graphing circles from features Mathematics II High School Math Khan Academy.mp3 |
I need to make this radius equal to 5. So let's see if I take that. So now the radius is equal to 2, 3, 4, and 5. There you go. Centered at 3, negative 2, radius of 5. Notice, go from the center to the actual circle, it's 5, no matter where you go. Let's do one more of these. | Graphing circles from features Mathematics II High School Math Khan Academy.mp3 |
There you go. Centered at 3, negative 2, radius of 5. Notice, go from the center to the actual circle, it's 5, no matter where you go. Let's do one more of these. Graph the circle which is centered at negative 4, 1 and which has the point 0, 4 on it. So once again, let's drag the center. So it's going to be negative 4, x is negative 4, y is 1, so that's the center. | Graphing circles from features Mathematics II High School Math Khan Academy.mp3 |
Let's do one more of these. Graph the circle which is centered at negative 4, 1 and which has the point 0, 4 on it. So once again, let's drag the center. So it's going to be negative 4, x is negative 4, y is 1, so that's the center. It has the point 0, 4 on it. So x is 0, y is 4. So I have to drag, I have to increase the radius of the circle. | Graphing circles from features Mathematics II High School Math Khan Academy.mp3 |
So it's going to be negative 4, x is negative 4, y is 1, so that's the center. It has the point 0, 4 on it. So x is 0, y is 4. So I have to drag, I have to increase the radius of the circle. Let's see, whoops, nope, I want to make sure I don't change the center. I want to increase the radius of the circle until it includes this point right over here, 0, 4. So I'm not there quite yet. | Graphing circles from features Mathematics II High School Math Khan Academy.mp3 |
So I have to drag, I have to increase the radius of the circle. Let's see, whoops, nope, I want to make sure I don't change the center. I want to increase the radius of the circle until it includes this point right over here, 0, 4. So I'm not there quite yet. There you go. I am now including the point 0, 4. And if we're curious what the radius is, we could just go along the x-axis. | Graphing circles from features Mathematics II High School Math Khan Academy.mp3 |
So I'm not there quite yet. There you go. I am now including the point 0, 4. And if we're curious what the radius is, we could just go along the x-axis. x equals negative 4 is the x-coordinate for the center and we see that this point, so this is negative 4, 1 and we see that 1, 1 is actually on the circle. So the distance here is, you go 4 and then another one, it's 5. So this has a radius of 5. | Graphing circles from features Mathematics II High School Math Khan Academy.mp3 |
So first of all, there's this idea of rigid transformations, which we've talked about in other videos. But just as a refresher, these are transformations that preserve distance between points. So for example, if I have points A and B, a rigid transformation could be something like a translation because after I've translated them, notice the distance between the points is still the same. It could be like that. It includes rotation. Let's say I rotated about point A as the center of rotation. That still would not change, that still would not change my distance between points A and B. | Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3 |
It could be like that. It includes rotation. Let's say I rotated about point A as the center of rotation. That still would not change, that still would not change my distance between points A and B. It could even be things like taking the mirror image. Once again, that's not going to change the distance between A and B. What's not a rigid transformation? | Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3 |
That still would not change, that still would not change my distance between points A and B. It could even be things like taking the mirror image. Once again, that's not going to change the distance between A and B. What's not a rigid transformation? Well, one thing you might imagine is dilating, scaling it up or down. That is going to change the distance. So rigid transformation are any transformations that preserve the distance between points. | Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3 |
What's not a rigid transformation? Well, one thing you might imagine is dilating, scaling it up or down. That is going to change the distance. So rigid transformation are any transformations that preserve the distance between points. Now another idea is congruence. And in the context of this video, we're going to be viewing the definition of congruence as two figures are congruent if and only if there exists a series of rigid transformations which will map one figure on to the other. You might see other definitions of congruence in your life, but this is the rigid transformation definition of congruence that we will use. | Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3 |
So rigid transformation are any transformations that preserve the distance between points. Now another idea is congruence. And in the context of this video, we're going to be viewing the definition of congruence as two figures are congruent if and only if there exists a series of rigid transformations which will map one figure on to the other. You might see other definitions of congruence in your life, but this is the rigid transformation definition of congruence that we will use. And we're going to use these two definitions to prove the following. To prove that saying two segments are congruent is equivalent to saying that they have the same length. So let me get some space here to do that in. | Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3 |
You might see other definitions of congruence in your life, but this is the rigid transformation definition of congruence that we will use. And we're going to use these two definitions to prove the following. To prove that saying two segments are congruent is equivalent to saying that they have the same length. So let me get some space here to do that in. So first let me prove that if segment AB is congruent to segment CD, then the length of segment AB, which we'll just denote as AB without the line over it, is equal to the length of segment CD. How do we do that? Well, the first thing to realize is if AB, if AB is congruent to CD, then AB can be mapped onto CD with rigid transformations, rigid transformations. | Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3 |
So let me get some space here to do that in. So first let me prove that if segment AB is congruent to segment CD, then the length of segment AB, which we'll just denote as AB without the line over it, is equal to the length of segment CD. How do we do that? Well, the first thing to realize is if AB, if AB is congruent to CD, then AB can be mapped onto CD with rigid transformations, rigid transformations. That comes out of the definition of congruence. And then we could say since the transformations are rigid, distance is preserved, preserved, and so that would imply that the distance between the points are going to be the same. AB, the distance between points AB or the length of segment AB is equal to the length of segment CD. | Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3 |
Well, the first thing to realize is if AB, if AB is congruent to CD, then AB can be mapped onto CD with rigid transformations, rigid transformations. That comes out of the definition of congruence. And then we could say since the transformations are rigid, distance is preserved, preserved, and so that would imply that the distance between the points are going to be the same. AB, the distance between points AB or the length of segment AB is equal to the length of segment CD. That might almost seem too intuitive for you, but that's all we're talking about. So now let's see if we can prove the other way. Let's see if we can prove that if the length of segment AB is equal to the length of segment CD, then segment AB is congruent to segment CD. | Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3 |
AB, the distance between points AB or the length of segment AB is equal to the length of segment CD. That might almost seem too intuitive for you, but that's all we're talking about. So now let's see if we can prove the other way. Let's see if we can prove that if the length of segment AB is equal to the length of segment CD, then segment AB is congruent to segment CD. And let me draw them right over here just to, so let's say I have segment AB right over there, and I'll draw another segment that has the same length. So maybe it looks something like this, and this is obviously hand-drawn. So then let's call this CD. | Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3 |
Let's see if we can prove that if the length of segment AB is equal to the length of segment CD, then segment AB is congruent to segment CD. And let me draw them right over here just to, so let's say I have segment AB right over there, and I'll draw another segment that has the same length. So maybe it looks something like this, and this is obviously hand-drawn. So then let's call this CD. So in order to prove this, I have to show, hey, if I have two segments with the same length, that there's always a set of rigid transformations that will map one segment onto the other, which means by definition they are congruent. So let me just construct those transformations. So my first rigid transformation that I could do is to translate, translate, and I'll underline the name of the transformation, segment AB so that point A is on top of point C, or A is mapped onto C. And you could see that there's always a translation to do that. | Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3 |
So then let's call this CD. So in order to prove this, I have to show, hey, if I have two segments with the same length, that there's always a set of rigid transformations that will map one segment onto the other, which means by definition they are congruent. So let me just construct those transformations. So my first rigid transformation that I could do is to translate, translate, and I'll underline the name of the transformation, segment AB so that point A is on top of point C, or A is mapped onto C. And you could see that there's always a translation to do that. It would be doing that. And of course we would translate, B would end up like that. And so after this translation, it's going to be A right over there. | Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3 |
So my first rigid transformation that I could do is to translate, translate, and I'll underline the name of the transformation, segment AB so that point A is on top of point C, or A is mapped onto C. And you could see that there's always a translation to do that. It would be doing that. And of course we would translate, B would end up like that. And so after this translation, it's going to be A right over there. A is going to be there. And then B is going to be right over there. Now the second step I would do is then rotate AB about A, so A is the center of rotation, so I'm gonna rotate it, so that point B lies on ray CD. | Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3 |
And so after this translation, it's going to be A right over there. A is going to be there. And then B is going to be right over there. Now the second step I would do is then rotate AB about A, so A is the center of rotation, so I'm gonna rotate it, so that point B lies on ray CD. Well, what does this transformation do? Well, since point A is the center of rotation, A is going to stay mapped on top of C from our first translation, but now B is rotated, so it sits on top of the ray that starts at C and goes through D and keeps going. And where will B be on that ray? | Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3 |
Now the second step I would do is then rotate AB about A, so A is the center of rotation, so I'm gonna rotate it, so that point B lies on ray CD. Well, what does this transformation do? Well, since point A is the center of rotation, A is going to stay mapped on top of C from our first translation, but now B is rotated, so it sits on top of the ray that starts at C and goes through D and keeps going. And where will B be on that ray? Well, since the distance between B and A is the same as the distance between D and C, and A and C are at the same point, and now B sits on that ray, B will now sit right on top of D because AB is equal to CD, B will be mapped onto D. And just like that, we've shown that if the segment lengths are equal, there is always a set of rigid transformations that will map one segment onto the other. Therefore, since A and B have been mapped onto C and D, we know that segment AB is congruent to segment CD, and we are done. We have proven what we set out to prove both ways. | Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3 |
And where will B be on that ray? Well, since the distance between B and A is the same as the distance between D and C, and A and C are at the same point, and now B sits on that ray, B will now sit right on top of D because AB is equal to CD, B will be mapped onto D. And just like that, we've shown that if the segment lengths are equal, there is always a set of rigid transformations that will map one segment onto the other. Therefore, since A and B have been mapped onto C and D, we know that segment AB is congruent to segment CD, and we are done. We have proven what we set out to prove both ways. | Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3 |
We have proven what we set out to prove both ways. | Segment congruence equivalent to having same length Congruence Geometry Khan Academy.mp3 |