In Exercises 33–42, use the center, vertices, and asymptotes to g...

Question

In Exercises 33–42, use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. (x−3)^2−4(y+3)^2=4

Accepted Answer

Hello, today we're gonna be graphing the equation of the hyperbola. Then finding the equations of the diagonal asymptotes and finding the coordinates of the ci. So what we are given is X plus one squared minus nine times the quantity Y minus two squared is equal to nine. Now, in order to get this into the form of an equation of a hyperbola, we need to go ahead and set the equation equal to one in order to do this, we're going to divide every term in the equation by nine. And that is going to allow us to simplify the equation to X plus one squared over nine minus Y minus two squared over one is equal to one. Now, there are a few things to note about this equation of the hyperbola. First, the X and Y quantities are given to us as binomials. And what that means is that this equation of a hyperbola is not located at the origin, the center is going to be defined by H comma K. And another thing to note is that the X term is the leading term on the left hand side of the equation. What this means is that the major axis for the equation of the hyperbola is going to be the X axis. So since we know that this is not located at the origin and the major axis of the X axis, the general form of the equation of the hyperbola is going to be X minus H squared over A squared minus Y minus K squared over B squared is equal to one. This means that we have an equation of the hyperbola where the hyperbola opens up to the left and to the right, the vertices are some distance of A away from the center of the hyperbola. And the height of the ascent total box is some distance B from the center of the hyperbola. So what we need to do is we need to use the general form of the equation of the hyperbola to get our AM B distances and to get the center of the hyperbola. So A squared is currently equal to nine and B squared is equal to one. So we can set A squared equal to nine and B squared equal to one. If we take the square root of both sides of both equations, we get A is equal to plus or minus three and B is equal to plus or minus one. So that means that the vertices are going to be a distance of three away from the center of the hyperbola. And the height of the total box is going to be a distance of one away from the center of the hyperbola. Now let's go ahead and find our center. If we take a look at the X quantity given to us, we are given X plus one quantity squared. This is not in the general form of X minus H squared, but it can be rewritten as X minus negative one squared. Now, this is in the general form of X minus H squared where H is equal to negative one. Now let's get our K value, we are given Y minus two quantity squared. This is in the general form of Y minus K squared. And this is where K is equal to positive two. So that means that the center of the hyperbola is given to us as negative one comma two. Now that we have the center of the hyperbole and the distance of the vertices, we can go ahead and create a rough sketch of the hyperbola. So the hyperbola is gonna be located at negative one comma two. Since this is a hyperbola that opens up to the right and to the left. That means the vertices are going to be a distance of three to the right and to the left of the center of the hyperbola. That means that the vertices are gonna be located at two comma two and negative four comma two. And the height of the ascent total box is one unit above and below the center of the hyperbole. This is going to be located at negative one, comma one, a negative one comma three. Now we can go ahead and connect these four dots and draw our as in total box. And we can even use this box to create a rough sketch of the diagonal asymptotes. Now, we can go ahead and draw the hyperbola with respect to the ascent totes and the shape of a hyperbola is roughly gonna look like this. So now that we have the shape of the hyperbola, the next thing we wanna do is we want to find the locations of the. Now the are going to be some distant c away from the center of the hyperbola. And for hyperbolas, we do have an equation that helps us find the C variable, we have C squared is equal to A squared plus B squared. Now, we know what the A squared and B squared values are going to be A squared is equal to nine and B squared is equal to one. So if we substituting these values, we get C squared is equal to nine plus one and nine plus one is going to give us 10. So we currently have C squared as is equal to 10. And if we take the square root of both sides of the equation, we get C is equal to the square root of 10. So what this means is that the are gonna be located a distance of square root of 10 to the right and to the left of the center of the hyperbola. That means we're going to be adding this attracting the value of square root of 10 from the X value of the center of the hyperbola. And that means that the side gonna be located at negative one plus square root of 10 comma two, a negative one minus square root of comma two. So these are gonna be the locations of the. Now, what we need to do is we need to find the equations of the diagonal ascent tode. Now again, recall that in the beginning that we said that this is an hyperbola not located at the origin. And the major axis is the X axis. So that means the general form of the equations of the diagonal asy totes is going to be Y is equal to plus or minus B over A times the quantity X minus H plus K. So we already know what our B N A values are going to be B is equal to one and A is equal to three. We also know that H is equal to negative one and K is equal to positive two. So if we plug this into the general form of the equations of the diagonal asymptotes, we end up getting two equations. The first one being that Y is equal to positive 1/3 times the quantity X minus negative one, which is X plus one. Plus K which is plus two. And we also get the negative version, Y is equal to negative 1/3 times X plus one plus two. So on the left hand side, we're going to distribute one third into X plus one that is going to give us Y is equal to 1/3, X plus 1/3 plus two. And on the right hand side, we're going to distribute negative one third into X plus one that it's going to give us Y is equal to negative 1/3, X minus 1/3 plus two. On the left hand side, if we add one third and two together, we get the value of 7/3. So the first equation is gonna be given to us as 1/3, X plus 7/3. And on the right hand side, if we subtract one third from two, that is going to give us the value of 5/3. So we get Y is equal to 1/3 X plus 5/3. And these are going to be the equations of the diagonal asymptotes. So just to recap, we found the locations of the, we found the equations of the Diago Asos and we found the rough shape of the equation of the hyperbola. With that being said, this is going to be the answer to this problem. So I hope this video helps you in understanding how to approach this problem and I'll go ahead and see you all in the next video.

Struggling with College Algebra?

In Exercises 33–42, use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. (x−3)^2−4(y+3)^2=4

Watch next

Hyperbolas at the Origin practice set