Hyperbolas at the Origin - Online Tutor, Practice Problems & Exam Prep

Hyperbolas are unique conic sections characterized by their equations, which differ from ellipses primarily by a minus sign. The vertices, located at a distance 'a' from the center, and foci, determined by the equation \(c^2 = a^2 + b^2\), are essential for graphing. Asymptotes, found using the values of 'a' and 'b', guide the shape of the hyperbola. Understanding these components allows for effective graphing and analysis of hyperbolas, reinforcing the connections between their geometric properties and algebraic representations.

Introduction to Hyperbolas

Hey, everyone, and welcome back. So up to this point, we have talked about circles, ellipses, and parabolas. And we're now going to move on to the hyperbola shape in this series on conic sections. In my opinion, the hyperbola is both the most unique and oftentimes the most difficult of all four shapes that we've covered, and the reason for this is because there's a lot to remember about hyperbolas. But don't sweat it, because over the course of this video and the next few videos on conic sections, I think you're going to find that hyperbolas actually correlate very similarly to a lot of concepts that we've already learned. So without further ado, let's get right into this.

Now the equation for hyperbolas is nearly identical to the equation for an ellipse. The only difference is there is a minus sign rather than a plus sign. So if we take a look at the horizontal hyperbola, for example, notice in this equation for the hyperbola, it's very similar to what we've learned previously about the ellipse. The only real difference is that we have a plus sign for the ellipse and a minus sign for the hyperbola. Now even though the equations for the ellipse and hyperbola are similar, the shapes are actually quite different. Because visually, a hyperbola appears as two parabolas that are facing away from each other.

Now something else that I'll mention, this ellipse that we see drawn on this graph, you're not gonna actually need to ever draw this ellipse when dealing with hyperbolas. This is just here so you can see the similarities between the ellipse equation and the hyperbola equation and what the a and b values really mean. Now looking at this horizontal hyperbola, notice that we can see what the a value is for the ellipse that we drew in here. The a value is 1, 2, 3 units long. And we've talked about in previous videos how a represents the semi-major axis of an ellipse, whereas b, which is 1, 2 units long, represents the semi-minor axis. Now when it comes to the hyperbola, this a value tells you the distance to get from the center of the hyperbola to either of the two curves. So as you can see, this is the shortest distance to get to this curve, and the shortest distance to get to that curve.

Now this b value is a little bit less intuitive when it comes to the hyperbola, because we know that this b value describes the semi-minor axis for the ellipse, but for the hyperbola, it actually serves a bit of a different purpose. It can oftentimes help in describing the height of the hyperbola, but it's also going to be very critical to use this b value when it comes to graphing the hyperbolas, which we'll talk about in later videos.

So this is the main idea behind the horizontal hyperbola, and this is what the equation would look like if we use this a and b value because, remember, you have to square both the terms in the denominator. But now let's take a look at the vertical hyperbola. Notice for the vertical hyperbola, we have an a value of 1, 2, 3, except the a value is now up and down to get to either of the curves for the hyperbola rather than left and right. And notice how the b value, which is 1, 2 units long, is now left and right as opposed to being up and down.

Now for the ellipse, we've talked about how when you have a vertical ellipse that the semi-major axis squared is going to be underneath the y value this time, and the semi-minor axis would be under the x value. We'll notice for the hyperbola, we have a similar situation, where we have the a squared underneath the y squared and the b squared underneath the x squared. So basically, all we did was take the x and the y and switch them in this version of the equation. So the equation for the vertical hyperbola is going to be: y2 9 - x2 4 = 1 . And, again, this is because you have to square the a and b value when you put them in this equation. Now there's one more big difference I wanna mention between the ellipse and the hyperbola, and that is the major axis a. Because when dealing with the ellipse, we're used to this a value always being the largest value that we see. Well, when it comes to the hyperbola, the a is actually the first value that you see rather than the largest. So for these equations in this example, we saw that a was the biggest number, but this a is not always gonna be biggest for the hyperbola. This a is simply always going to come first in the equation.

Now you may have felt like this was a lot of information, and it was. But let's actually try an example, see if we can really solidify this altogether. So in this example, we have some equations for hyperbolas, and we're asked to match each equation with the corresponding graph. Now we're gonna start by looking at this first equation, equation a. Now one of the first things I notice is that we have the y squared in front. And remember, this front term that you see is always going to be the a squared term. So you can see that a squared is equal to 16, this number here, which means that a is the square root of 16, which is 4. So what I see here is that we have a y squared and that our a value is 4. Since we have the y squared in front that means we're going to have a vertical hyperbola, and our a value is going to be 4. Now if I look at these three graphs, the only vertical hyperbola I see is this one on the far right side. So if I take a look at graph number 3 here, I can see that our a value is 4. So if I start from the center, I can go 1, 2, 3, 4 units up, and 1, 2, 3, 4 units down. This does check out, which means that graph number 3 is going to match with equation a. But now let's take a look at equation b. For equation b, I can see that our a squared value which always comes first is going to be 4, which means a is gonna be the square root of 4 which is 2.

So you can see that we have an x squared up front which means we're going to have a horizontal hyperbola, and I can see that our a value is 2. Now if I go ahead and start from the center of this hyperbola on graph number 2, I can go 1, 2 units to the right and 1, 2 units to the left. Well, this is a problem, because notice that our a values go outside of the curve. So this is not the correct hyperbola that we're looking for. But now let's try graph number 1. If we draw out our a value, we can start in the center and go 1, 2 units to the right and 1, 2 units to the left. And this lands us directly where our curve is, so this graph does check out. Now just by process of elimination, it's clear to see that equation c is going to match with graph 2. But let's actually see if we can understand why. So I can see that the first term that we have here is 1, and this is underneath x squared. X squared means we're going to have a hyperbola that opens left and right, and this one means that our a squared value is equal to 1, meaning a is going to be the square root of 1, which is just 1. So if I start in the center of this hyperbola that we have here, I can go 1 unit to the right and 1 unit to the left. This is going to be our a values, and that does match with the curve that we have here. So we can see that equation c matches with graph number 2.

So this is the main idea behind the graphing and equation of hyperbolas. Hope you found this video helpful, and thanks for watching.

Given the hyperbola $\frac{x^2}{25}-\frac{y^2}{9}=1$, find the length of the $a$-axis and the $b$-axis.

• A. $a=25,b=9$
• B. $a=9,b=25$
• C. $a=5,b=3$
• D. $a=3,b=5$

Given the hyperbola $x^2-\frac{y^2}{4}=1$, find the length of the $a$-axis and $b$-axis.

• A. $a=1,b=4$
• B. $a=4,b=1$
• C. $a=1,b=2$
• D. $a=2,b=1$

Given the hyperbola $\frac{y^2}{100}-\frac{x^2}{139}=1$, find the length of the $a$-axis and the $b$-axis.

• A. $a=100,b=139$
• B. $a=139,b=100$
• C. $a=\sqrt{139},b=10$
• D. $a=10,b=\sqrt{139}$

Foci and Vertices of Hyperbolas

Hey everyone and welcome back. So up to this point we've been talking about the various shapes that you get from conic sections, and in the last video, we looked at hyperbolas. Now in this video, what we're going to do is see how we can find the vertices and foci of a hyperbola. Now I will warn you that this process is pretty tedious because there are some new equations that you'll need to know in order to solve these types of problems, and you are going to be asked to solve these at some point in this course. But don't sweat it because in this video, we're going to be going over some examples and scenarios that I think will make this process seem a lot clearer. And I'll also mention that I think you'll find there are a lot of similarities in the problem-solving with hyperbolas that we did with ellipses in previous videos. So without further ado, let's get right into this.

Just like an ellipse, every hyperbola has 2 vertices and 2 foci, and they are on the major axis. Now the vertices are the points of the hyperbola that are closest to the center. And this distance from the center to either vertex is a distance of \(a\). Now to understand this a bit better, let's actually take a look at this example we have down here. Where we're asked to find the vertices and foci of the hyperbola in the graph. Now in order to find the vertices, we said they are a distance \(a\) from the center, and we can see the center is at the origin here \(0\). Now recall that \(a^2\) is going to be the first thing that we see in our denominator assuming we have the standard form of \(x^2/a^2 - y^2/b^2 = 1\). And we can see that we do have the standard form here, that means \(a^2\) is going to be equal to this first thing in the denominator which is \(4\). That means to calculate \(a\) we just need to take the square root of \(4\) which is \(2\). So if \(a = 2\), then we can start here at the center of our hyperbola and we can go 2 units to the left and 2 units to the right starting from the center. This puts our vertices at \((-2, 0)\) and \((2, 0)\), and that is how you can calculate the vertices of a hyperbola.

Now you also will need to know how to find the foci. And the thing that makes the foci or the focus points unique is that these are going to be a situation where for any point on the hyperbola the difference of the distances from each focus point to any point of the hyperbola is going to be constant. Now to make sense of this, let's say that we have this hyperbola down here and let's say that the foci end up right about there and right about here. If these are the 2 foci, what you can do is find a point, any point somewhere on the curve of the hyperbola. And if you take this distance and you subtract off that distance, the number that you get will be constant for any point on the hyperbola. So let's say this distance is \(5\) and that distance is \(2\). \(5 - 2 = 3\). And that means any point that you look at anywhere on the hyperbola, the two distances from the foci to there, if you find their difference, it will always come out to \(3\). So that's what makes the foci unique.

Now in order to calculate the foci, you will need the distance \(c\). And \(c\) can be calculated from this equation c2=a2+b2. Now notice that this equation looks a lot like the equation we had for the ellipse where c2=a2-b2. The only difference is for the foci of a hyperbola, c2=a2+b2. Now you might notice at this point there are a lot of similarities for the equations between the hyperbola and the ellipse. The main differences are the fact that we have a plus sign here rather than a minus sign for the hyperbola, and then we have a minus sign rather than a plus sign for the standard equation of a hyperbola. So just keep in mind that there are going to be differences in the signs and how you use the equations generally, but a lot of the problem-solving ends up looking pretty similar.

Now let's actually see if using this equation we can calculate the foci on this hyperbola. So c2=a2+b2 And recall that \(b^2\) is going to be equal to the second thing we see in the denominator in the standard form. So that means that \(b^2\) is equal to \(9\), meaning that \(b\) is going to be the square root of \(9\), which is \(3\). So in order to calculate \(c\), what we can do is we can recognize that \(a^2\) is going to be \(2^2\), which is \(4\), and \(b^2\) is going to be \(3^2\) which is \(9\). So \(c^2\) is equal to \(4 + 9\) which is \(13\). If I take the square root on both sides of this equation, we'll get that \(c\) is equal to the square root of \(13\). And the square root of \(13\) is approximately equal to \(3.6\). So what that means is that if we start at the center of our hyperbola here I can go approximately \(3.6\) units to the right which will get me to this point at \(\sqrt{13}\), 0 and I can go approximately \(3.6\) units to the left getting me at this point which is \(-\sqrt{13}\), 0. And this is how you can find the vertices as well as the foci for a hyperbola.

Now one more thing that I want to mention is how the vertices and foci will look different depending on the hyperbola's orientation. Because we just looked at an example of a horizontal hyperbola and notice how the vertices ended up on the x-axis at \(a\), 0 and \(-a\), 0 and notice how the foci also ended up on the x-axis at \(c\), 0 and \(-c\), 0. So what we can conclude is that when we have a horizontal hyperbola, the vertices and foci end up on the x-axis. Now if we instead had a vertical hyperbola, the vertices and foci would actually end up on the y-axis. So you'd end up seeing the vertices end up around here and there at \(0, a\), \(0, -a\), and you find the foci would end up just outside of these two curves just like we had with the horizontal hyperbola, except these would be at \(0, c\) and \(0, -c\). So when it's a vertical hyperbola, the vertices and foci are on the y-axis. As for a horizontal hyperbola, they're on the x-axis. So that is how you can find the vertices and foci of a hyperbola. Hope you found this video helpful. Thanks for watching.

Determine the vertices and foci of the hyperbola $\frac{y^2}{4}-x^2=1$.

• A. Vertices: $\left(2,0\right),\left(-2,0\right)$Foci: $\left(\sqrt5,0\right),\left(-\sqrt5,0\right)$
• B. Vertices: $\left(0,2\right),\left(0,-2\right)$Foci: $\left(0,\sqrt5\right),\left(0,-\sqrt5\right)$
• C. Vertices: $\left(1,0\right),\left(-1,0\right)$Foci: $\left(5,0\right),\left(-5,0\right)$
• D. Vertices: $\left(0,1\right),\left(0,-1\right)$Foci: $\left(0,5\right),\left(0,-5\right)$

Find the equation for a hyperbola with a center at $(0,0)$, focus at $(0,-6)$ and vertex at $(0,4)$ .

• A. $\frac{y^2}{16}-\frac{x^2}{20}=1$
• B. $\frac{y^2}{20}-\frac{x^2}{16}=1$
• C. $\frac{y^2}{4}-\frac{x^2}{\sqrt{20}}=1$
• D. $\frac{y^2}{\sqrt{20}}-\frac{x^2}{4}=1$

Asymptotes of Hyperbolas

Hey, everyone, and welcome back. So in the last video, we talked about how you can find the foci and vertices for hyperbolas. Continuing on this shape, we're going to now discuss how you can find the asymptotes of hyperbolas. This might sound a bit complicated because, up to this point, we haven't really dealt with asymptotes for any of these shapes. I will warn you that the asymptotes of a hyperbola can be a bit complicated when you first learn it. But what we're going to learn in this video is that even though the process is a little bit tedious, there is a pattern that forms when you deal with the asymptotes of a hyperbola. This pattern allows us to find a general equation that makes solving the asymptotes of a hyperbola super quick and straightforward. So, without further ado, let's get right into this.

The asymptotes are required when graphing hyperbolas. To find the asymptotes, you can use the a and b values. Recall that a is the major axis and b is the minor axis for a hyperbola. These a and b values form a box where you can take lines and draw them through the corners of this box. These lines that you draw are going to be the asymptotes of your hyperbola. Let's take a look at this example we have down here. We're asked to find and draw the asymptotes of the given hyperbola. First, I'm going to see if I can find my a and b values. Recall that a² is equal to the first thing we see in the denominator, assuming that we have standard form, which we do. I can see that a² is going to be 9. To find a, I just need to take the square root on both sides, giving me that a is the square root of 9, which is 3. Now, to find b, recall that b² is equal to the second thing we see in the denominator. In this case, it's 4. So to find b, I just take the square root on both sides here, giving me that b is the square root of 4, which is 2.

So, I have our a values here and there, which also correspond with the vertices points, and I can see that our b points are going to be to the left 2 units and to the right 2 units. From here, we can now draw this box that connects these four points together. Once you have this box drawn, you can draw the asymptotes through the corners of the box. One line is going to go through the corners, and it's going to cross through the center, assuming that your hyperbola is at the origin like we have here. This is the first asymptote. The second asymptote is going to go through the other direction. To define the equation for each of the asymptotes, we recognize that we have two lines that we formed. The equation for a line is y = m x + b , where m is the slope of the line and b is the y-intercept. Both of these lines cross through the center, meaning that our y-intercept is going to be 0 since it's right at the origin of the graph. Our slope is going to be rise over run. For this line that slopes up, our rise is going to be 3 units up, divided by the run, which I can see we have to run 2 units to the right. So our slope is 3/2, and our y-intercept is 0. Plugging everything into this equation, what I'll have is y = 3 2 x . Now, for the other line, we can find the equation for this asymptote pretty quickly because this is the same line, it just slopes down now. So all we're going to do is change this slope to negative, meaning that we're going to have y = − 3 2 x . So, this is how you can find the equations for the asymptotes of your hyperbola.

It may seem like it was a long and tedious process because we first had to draw out this graph, find the a and b points, and then the lines and then the line equation, but you're going to find that there's actually a pattern that forms if you look at the equations that you got. Because notice that our rise was 3, well, that's just associated with what we have for a. And notice that our run was 2. Well, that's just b. So it turns out that when dealing with vertical hyperbolas, the equation for the asymptotes are always going to be, assuming you're at the origin, they're always just going to be plus or minus a/b. And we can see here that that's exactly what we got. We got a positive 3/2 x and a negative 3/2 x.

So that is how you can quickly find the asymptotes for a vertical hyperbola. Thanks for watching, and let me know if you have any questions.

Find the equations for the asymptotes of the hyperbola $\frac{x^2}{64}-\frac{y^2}{100}=1$.

• A. $y=\pm \frac{4}{5}x$
• B. $y=\pm \frac{5}{4}x$
• C. $y=\pm \frac{16}{25}x$
• D. $y=\pm \frac{25}{16}x$

Find the equations for the asymptotes of the hyperbola $\frac{y^2}{16}-\frac{x^2}{9}=1$.

• A. $y=\pm \frac{9}{16}x$
• B. $y=\pm \frac{16}{9}x$
• C. $y=\pm \frac{3}{4}x$
• D. $y=\pm \frac{4}{3}x$

Graph Hyperbolas at the Origin

Hey, everyone. Welcome back. So, in the previous video, we talked about finding the asymptotes of a hyperbola. And in this video, we're going to see if we can take all this knowledge that we've learned about hyperbolas and combine them into graphing hyperbolas from just the equation. Now, this process has a lot of steps to it, but if you've watched the previous videos up to this point, hopefully, all these steps will make a lot of sense. And don't worry because I'm going to be explaining them all in this video. So, without further ado, let's get right into this.

Now here we have an example where we're asked to graph the hyperbola and identify the foci given this equation. Our first step should be to determine whether the hyperbola is horizontal or vertical, and I can do that by just looking at the equation that we have. Recall that the first term that you see in the denominator is the a² term, and this a² term is underneath x². Since we see an x that shows up first, that means this hyperbola will be oriented here on the x-axis. So because of this, we could say that we are dealing with a horizontal hyperbola.

Now, our next step should be to identify the vertices. And because we have a horizontal hyperbola, we only are going to care about the x-coordinate for the vertices here. So we're going to use this version of the coordinates. Now to find out what a is, well, a² is equal to 9. So if a² is equal to 9, that means a is going to be the square root of 9, and the square root of 9 is equal to 3. So our vertices are going to be at 3 and at negative 3.

Now, step 3 here tells us that we need to find our b values, and to find our b values, we need to look at this other value that we have in the denominator. This corresponds with b, and we see that b² is equal to 64, which means b is going to be the square root of 64. The square root of 64 turns out to be 8. So that means that our b values are going to now follow this pattern, and by the way, it's this pattern because since we had our vertices on the horizontal axis, our b values are going to be on the vertical axis. So we're going to have 0, 8, and 0, -8.

Now, our fourth step is going to be to find the asymptotes, and this is going to be a two-step process. Step a is going to be to draw a box through the vertices and b points. So basically, with all the points that we found so far, I can connect these all with a rectangular box. And step b for drawing the asymptotes is to draw two lines through the corners of the box. So if I draw lines through the corners, one line is going to look like this, and another line is going to look like that. So these are the two lines that we need to draw, which are also the asymptotes for our hyperbola.

And our last step, which is step 5, is going to be to draw branches at the vertices approaching the asymptotes, which are these lines. So if I start at the vertices here, I can draw a curve that looks like this, and that looks like that. Notice how it approaches these lines that we see. And another curve is going to be drawn like this and like that as well. So this is going to be what the hyperbola looks like. So now that we have our two branches or curves, we have successfully graphed this hyperbola.

And as a last step, we're asked to also find the foci of this hyperbola. Well, to do this, recall that we need to use this equation, which is c² ≡ a² + b². Our c² is going to equal a², and we said up here that a² is equal to 9, so it's going to be 9 + b², which we said was equal to 64. It's gonna be 9 + 64. 9 + 64 comes out to 73, which I'm going to write over here. So we're going to have the c² is equal to 73, and then we're going to take the square root on both sides of this equation, giving us that c is equal to the square root of 73, and the square root of 73 is approximately equal to 8.54. This is just the approximate value for the square root of 73, but to put in this value here, we can say that our foci are going to be on the x-axis because it's a horizontal hyperbola. So we're going to have the square root of 73 and 0, and then we're going to have negative square root of 73 and 0 as our foci. And since we said that the c value is approximately equal to 8.54, and it's on the x-axis, that means one of our foci is going to be right about there, and another foci is going to be right about here. So these are the foci for the hyperbola, and that is how you can graph hyperbolas if you're only given the equation. So I hope you found this video helpful. Thanks for watching, and let's move on.

Graphing Hyperbolas at the Origin Example 1

Welcome back, everyone. Let's give this example a try. So in this example, we are given this equation, and we're asked to graph the following hyperbola. Now, our first step is going to be to determine the orientation of the hyperbola. And I can see that our x² comes first. And since the x comes first, that means we're going to be oriented on the x-axis. So, we are dealing with a horizontal hyperbola. Now, the next thing we need to do is find the vertices. And since we know that it's going to be along the x-axis, we only care about this x coordinate, so we're going to use these coordinates right here. Now first off, I need to find a and recall that a² is associated with the first thing we see in the denominator or 16. I can take the square root on both sides to get a by itself, which a will be the square root of 16, which is 4. So that means that our vertices are going to be at positive 4 and at negative 4. And if we go over to our graph, the vertices will be right about here and right about there. So that's how we can draw the vertices.

Now the next thing we need to do is find the b points, and for the b points, now these are going to be opposite the direction of the vertices since the vertices were left and right the b points are going to be up and down. And so we're going to use this coordinate, and we need to calculate b. So we can see that b² is going to be associated with the second number in the denominator which is 20. And this is kind of interesting. So if we take the square root on both sides, the square root of 20 doesn't come out to a nice number like this one did. So we're going to end up with b is equal to the square root of 20. Now, you could reduce this down to 2 times the square root of 5, but for these problems, I'm just gonna keep them in this whole form because I think it's gonna be a little easier to deal with. So b is going to be the square root of 20, and the square root of 20, on a calculator, comes out to around 4.47. So this is about 4.47, and so what that means is our b points are going to be up to right about there, about halfway between 4.5, and then the negative one is gonna be down here, right around between negative 5 and negative 4. So the b points are going to be at square root 20 0, and negative square root 20 0.

Now the next step is going to be to find the asymptotes, and to find the asymptotes we first need to draw a box. Now, I'll draw a box through the vertices and through the b points, which is going to look something like this. So this is what the box is going to look like, and then what we need to do from here is draw lines through the corners of the box. So one of the lines is gonna go through this corner, and another line is gonna go through that corner. So we now have the asymptotes drawn for this hyperbola, and that was step 4. And our 5th step is going to be to draw the branches of the vertices that approach the asymptotes. So one of these branches is going to start the vertices and go like this, and it's going to approach the asymptote that we drew, and it's going to extend down here as well. And then the other branch that we need to draw is going to start at this vertex and go up like this, and start to touch the or approach touching the asymptote, and it's going to go down as well there. So this right here is going to be the graph of our hyperbola, and that was step 5.

Now, we also need to find the foci. And to find the foci, well, since we're dealing with a horizontal hyperbola, the foci are going to be to the left and to the right. And what we need to do is use the equation c² is equal to a² + b², and a² + b², well, we determined that a² is what's under the denominator first, which is 16, and b² is what's under this denominator, which is 20. So I have that c² is 16 plus 20 which is 36, and then what I need to do is take the square root on both sides which will give a c. I'll go ahead and write c over on this side, c is going to be the square root of 36, which is 6. So, that means that our foci are going to be at 6 and negative 6, and we're gonna have to go a little bit off our graph to do this, but if we imagine that this line extends back here to negative 6, one of the vertices would be right about there, and if this graph were to extend to the right at positive 6, which let's say it's right about here, then that's where our other vertices is going to be. So these are going to be the 2 vertices as well as the graph, and that is going to be the answer for this example. So, hope you found this video helpful. Let me know if you have any questions.