Hyperbolas NOT at the Origin - Video Tutorials & Practice Problems

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Graph Hyperbolas NOT at the Origin

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Hey, everyone and welcome back. So up to this point, we've been talking about the four main types of conic sections and the various shapes you can get the shapes that we've covered have been the circle ellipse and Parabola. And in recent videos, we've been talking about this last shape, which is the hyperbola. Now, all the hyperbola we've dealt with so far have been centered at the origin of the graph. But the question becomes what happens if we deal with the hyperbola that is centered at some new location, that's not at the origin. Now, unfortunately, when this happens, the equation and graph are going to change and some of these problems can be a bit tedious to solve because of that, but don't sweat it because in this video, we're going to be going over some examples. And I think you're going to find that the way that the equation graph changes is very similar to what we've already learned with how the ellipse changes as well as what we've learned with function transformation. So without further ado let's get right into things. Now, the equation for a hyperbola that is centered at the origin looks like this, this is the Standard Equation. And if you have a hyperbola that is not of the origin, meaning it's shifted to some new location, then the equation looks like that. Now, you might notice that these equations are basically the same, the only real difference is that we is that we now have an H and A K that's being subtracted from the X and Y respectively. And this H and K mean the exact same things that they've always meant with transformations or with the ellipse shift where H is the horizontal position in is the vertical position with respect to where the origin is now to make sure this equation makes sense. Let's actually try an example where we have to graph a hyperbola that is not at the origin. So in this example, we are given the equation of a hyperbola and we're asked to graph it over here. Now, the first thing we should do is determine whether we have a horizontal or vertical hyperbola. Now, by looking at this equation, I noticed that the first thing that we see is a why? Because the second thing we have this minus sign here and so is always the positive minus whatever this term is. And since the Y shows up first here, that means that we're going to have a vertical hyperbola because it's like it's not necessarily on the Y axis, but it's vertically oriented. So this means for the first step, this is a vertical hyperbola. Now, our next step is going to be to find the center HK and to do that, we can use the equation we just discussed, notice that K is what is subtracted from the Y and here we have one, subtracted from the Y. So K is equal to one. And notice that H is subtracted from the X. And in this equation, the H would be and going down to our graph horizontally, we would be over to this two and vertically, we would be between zero and two, which is right about there. So this would be the center at 21. Now, our third step is going to be to find the vertices. And since we're dealing with a vertical hyperbola, we're going to use these coordinates. Now, what I first I'm going to do is find this a value and I can do this by looking at whatever this first positive term is and recognizing that that is equal to A squared. So A squared is equal nine. And then if I take the square root on both sides of this equation, we'll get the A is the square root of nine, which is three. Now, what I can see here from these coordinates is that H is going to stay constant for both of these. So H is two. So we're gonna have two for both of these. And then for K, we're going to add a and subtract a well, we said that K is one and then if I go ahead and add a to it, we're gonna have one plus three, which is four. And what I can do is take our one and subtract three, which will give us negative two. So our coordinates are going to be over at 24 and their coordinates are going to be at two negative two, which is down there. Now, our next step asks us to find the B points and since the vertices were up and down, our B points are going to be to the left and right. So what I can do is recognize if we're just going to the left and right, then the vertical position one is going to stay constant, but the horizontal position is going to change. So we have H plus B and H minus B. Now, what I need to do here is figure out what B is and B is going to be the second term because B squared is equal to whatever the second number is, what, where the minus signs in front of it and that's 16. And what I can do is take the square root on both sides to get that B is the square root of 16, which is four. Now, what I can do is take our H which is two and add four to it and that's going to give us positive six and then I can take our H and subtract four from it and 42 minus four gives us negative two. So the B points are going to be at 61, which is right about over here. And then it's also going to be at negative 21, which is back there. Now, our next step is going to be to find the asymptotes. And in order to find the asymptotes, what we first need to do is draw a box that connects these four points. So if I go ahead and draw this box, this is going to help us in drawing the asymptotes. And now to find the asymptotes, we need to draw lines through the corners of the box. We're gonna specifically draw these dotted lines, one line going to go through this corner and then the other line is going to go through the other corner of the box like this. So these are the two asymptotes. And our last step is going to be to draw branches at the vertices that approach the asymptotes. So the first branch is going to look something like this where it starts at the vertices or the vertex point. And then it approaches these asymptotes, we just drew and the other branch is going to be below this box, which is going to look something like that. So these are gonna be the two branches for this hyperbola. And now what we also can do now that we have our hyperbola graft is we can find the foi and again, we're dealing with a vertical hyperbola. So the fori coordinates are going to look like that. And for the foi notice that our H is going to stay the same. So H is going to be two for both of these. But RK we need to add and subtract C and recall for C. Well, C squared is equal to A squared plus B squared. We already discussed how A squared is going to be this first thing that we see in the denominator, which is nine. And then our B squared is the second thing that we see which is 16. And so we'll have the C squared is nine plus 16. And I'll write this over here. So C squared is 25 that's nine plus 16. And if we take the square root on both sides, we'll get that C is equal to the square root of 25 which is five. So what I can do is add five to K and we said that K was one. So five plus one will give us six and then we can go ahead and take RK value and subtract five and that's going to give us negative four. So the P I on our graph one is going to be at two comma six which here's two and then we go up to six and the other P I or focus point is going to be at two comma negative four. And if we go down to two and then negative four. That's going to be our other P I right there. So that is how you can find the F I as well as graph, the hyperbola. Hope you found this video helpful. Thanks for watching.

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Problem

Problem

Describe the hyperbola $\frac{\left(x+2\right)^2}{9}-\frac{\left(y-4\right)^2}{16}=1$.

A

This is a vertical hyperbola centered at $\left(-2,4\right)$ with vertices at $\left(4,2\right),\left(4,-6\right)$ and foci at $\left(4,4\right),\left(4,-8\right)$.

B

This is a vertical hyperbola centered at $\left(2,-4\right)$ with vertices at $\left(4,1\right),\left(4,-5\right)$ and foci at $\left(4,3\right),\left(4,-7\right)$.

C

This is a horizontal hyperbola centered at $\left(-2,4\right)$ with vertices at $\left(2,4\right),\left(-6,4\right)$ and foci at $\left(4,4\right),\left(-8,4\right)$.

D

This is a horizontal hyperbola centered at $\left(-2,4\right)$ with vertices at $\left(1,4\right),\left(-5,4\right)$ and foci at $\left(3,4\right),\left(-7,4\right)$.

3

Problem

Problem

Describe the hyperbola ${y}^{2}-\frac{{(x-1)}^{2}}{4}=1$.

A

This is a vertical hyperbola centered at $(1,0)$ with vertices at $(1,1),(1,-1)$ and foci at $(1,\sqrt{5}),(1,-\sqrt{5})$.

B

This is a vertical hyperbola centered at $(1,0)$ with vertices at $(1,2),(1,-2)$ and foci at $(1,1),(1,-1)$.

C

This is a horizontal hyperbola centered at $(-1,0)$ with vertices at $(0,0),(-2,0)$ and foci at $(\sqrt{5}-1,0),(-\sqrt{5}-1,0)$.

D

This is a horizontal hyperbola centered at $(1,0)$ with vertices at $(0,0),(-2,0)$ and foci at $(1,\sqrt{5}),(1,-\sqrt{5})$.

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