Hyperbolas NOT at the Origin - Online Tutor, Practice Problems & Exam Prep

Hyperbolas can be centered at points other than the origin, altering their equations. The standard form for a hyperbola centered at (h, k) is ${\mathrm{(x-h)}}^2/b^2-{\mathrm{(y-k)}}^2/a^2=1$. To graph, identify the center, vertices, and asymptotes, and calculate the foci using $c^2=a^2+b^2$. Understanding these transformations is crucial for graphing hyperbolas accurately.

Graph Hyperbolas NOT at the Origin

Hey, everyone, and welcome back. So up to this point, we've been talking about the 4 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 hyperbolas 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 a 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 at 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 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 and \( k \) 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 \( y \). Because the second thing, we have this minus sign here, and so it's 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, \( (h, k) \). 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 1. And notice that \( h \) is subtracted from the \( x \). And in this equation, the \( h \) would be 2. And going down to our graph, horizontally, we'd be over to this 2. And vertically, we would be between 0 and 2, which is right about there. So, this would be the center at (2, 1).

Now our 3rd 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 am gonna 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 \( a^2 \). So \( a^2 \) is gonna equal 9 and then if I take the square root on both sides of this equation, we'll get that \( a \) is the square root of 9 which is 3. Now what I can see here from these coordinates is that \( h \) is gonna stay constant for both of these, so \( h \) is 2. So we're gonna have 2 for both of these. And then for \( k \), we're going to add \( a \) and subtract \( a \). Well, we said that \( k \) is 1, and then if I go ahead and add \( a \) to it, we're gonna have 1 plus 3, which is 4, and what I can do is take our 1 and subtract 3, which will give us -2. So our coordinates are going to be over at (2, 4), and our coordinates are going to be at (2, -2), which is down there.

Now our next step is going to be 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 1 is going to stay constant. But the horizontal position is going to change. So we have \( h+b \), and \( h-b \). Now what I need to do here is figure out what \( b \) is, and \( b \) is going to be the second term because \( b^2 \) is equal to whatever the second number is, what were 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 \( b \) is the square root of 16, which is 4. Now what I can do is take our \( h \), which is 2, and add 4 to it, and that's going to give us positive 6. And then I can take our \( h \) and subtract 4 from it, and \( 2-4 \) gives us -2. So the \( b \) points are going to be at (6, 1), which is right about over here, and then it's also going to be at (-2, 1), 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 going to specifically draw these dotted lines. One line is 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 2 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 gonna look something like that. So these are going to be the 2 branches for this hyperbola.

And now what we also can do, now that we have our hyperbola graphed, is we can find the foci. And again, we're dealing with a vertical hyperbola, so the foci coordinates are going to look like that. And for the foci notice that our \( h \) is gonna stay the same, so \( h \) is gonna be 2 for both of these, but our \( k \) we need to add and subtract \( c \). And recall for \( c \), well \( c^2 \) is equal to \( a^2 \)+ \( b^2 \). And we already discussed how \( a^2 \) is gonna be this first thing that we see in the denominator, which is 9, and then our \( b^2 \) is the second thing that we see, which is 16. And so we'll have \( c^2 \) is 25, that's 9 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 5. So what I can do is add 5 to \( k \), and we said that \( k \) was 1. So 5 plus 1 will give us 6, and then we can go ahead and take our \( k \) value and subtract 5, and that's going to give us -4. So the foci on our graph, one's gonna be at (2, 6), which here's 2 and then we go up to 6. And the other foci or focus point is going to be at (2, -4). If we go down to 2 and then -4, that's going to be our other foci right there. So that is how you can find the foci as well as graph the hyperbola. I hope you found this video helpful. Thanks for watching.

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 $\left(3,4\right),\left(-7,4\right)$.

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 $(1,\sqrt{5}),(1,-\sqrt{5})$.

Here’s what students ask on this topic:

What is the standard form of a hyperbola not centered at the origin?

The standard form of a hyperbola not centered at the origin is given by the equation:

$\frac{{(x-h)}^2}{b^2}-\frac{{(y-k)}^2}{a^2}=1$

Here, (h, k) is the center of the hyperbola, a is the distance from the center to the vertices along the transverse axis, and b is the distance from the center to the co-vertices along the conjugate axis.

How do you find the vertices of a hyperbola not centered at the origin?

To find the vertices of a hyperbola not centered at the origin, first identify the center (h, k) from the equation. For a horizontal hyperbola, the vertices are located at (h ± a, k). For a vertical hyperbola, the vertices are at (h, k ± a). Here, a is the distance from the center to each vertex, which can be found from the equation where a² is the denominator of the positive term.

How do you determine the asymptotes of a hyperbola not centered at the origin?

The asymptotes of a hyperbola not centered at the origin can be found using the slopes derived from the equation. For a horizontal hyperbola, the asymptotes are given by the equations:

$y=k\pm \frac{a}{b}(x-h)$

For a vertical hyperbola, the asymptotes are:

$y=k\pm \frac{b}{a}(x-h)$

Here, (h, k) is the center, and a and b are the distances from the center to the vertices and co-vertices, respectively.

How do you find the foci of a hyperbola not centered at the origin?

To find the foci of a hyperbola not centered at the origin, first calculate c using the equation:

$c^2=a^2+b^2$

Then, for a horizontal hyperbola, the foci are located at (h ± c, k). For a vertical hyperbola, the foci are at (h, k ± c). Here, (h, k) is the center, a is the distance from the center to the vertices, and b is the distance from the center to the co-vertices.

How do you graph a hyperbola not centered at the origin?

To graph a hyperbola not centered at the origin, follow these steps:

1. Identify the center (h, k) from the equation.
2. Determine whether the hyperbola is horizontal or vertical by checking which term is positive.
3. Find the vertices using the distance a from the center.
4. Calculate the co-vertices using the distance b from the center.
5. Draw a rectangle using the vertices and co-vertices.
6. Draw the asymptotes by extending lines through the diagonals of the rectangle.
7. Sketch the hyperbola branches approaching the asymptotes.
8. Find the foci using c, where c² = a² + b², and plot them.