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)^2/25−y^2/16=1

Accepted Answer

Hello, today we're gonna be graphing the equation of the hyperbola. Then we're going to find the equations of the asy totes and the coordinates of the foci. So what we are given is X minus one squared over minus Y squared over nine is equal to one. Now, one thing to note is that the equation of the hyperbole is given to us in standard form. However, the X value is given to us as a binomial. So what this means is that the hyperbola is not located at the origin and the center is going to be defined as H comma K. Furthermore, the X variable is the leading variable on the left hand side of the equation. And that means that the major axis for the hyperbola is going to be the X axis. So with that being said, the general form of the equation of the hyperbole is given to us as X minus H squared over A squared minus Y minus K squared over B squared is equal to one. This is an equation of a hyperbola that opens up to the left and to the right. And this is where the vertices are defined as some distance of A away from the center of the hyperbola. And the height of the as in total box is some distance B from the center of the hyperbola. So let's go ahead and use the general form of the equation of the hyperbola to get the AM B values and to get the center of the hyperbola. So 36 is in the A squared spot and nine is in the B squared spot. We can go ahead and set A squared equal to B squared equal to nine. And if we take the square root of both of the equations, we get A is equal to plus or minus six and B is equal to plus or minus three. Now let's go ahead and get the center of the hyperbola. Let's go ahead and take a look at the X variable. We are given X minus one squared. This is in the standard form of X minus H squared. And this is where H is equal to one. Now let's take a look at the Y variable. We are just given Y squared, but this is the equivalent of Y minus zero squared. This is the standard form of Y minus K squared and this is where K is equal to zero. So that means the center of the hyperbola is given to us as one comma zero. Now that we know the center and the A and B values, let's go ahead and draw the general shape of this equation of the hyperbola. So again, we know that the center is located at one comma zero. The vertices are gonna be located a distance of six to the left and to the right of the hyperbola. So the vertices are gonna be located at seven comma zero and at negative five comma zero. And the height of the in total box is gonna be three units above and below the center of the hyperbola. So the height is going to be defined as one comma three and one comma negative three. Now we can connect these four dots and draw the as in total box. We can also use this box to create a rough sketch of the diagonal asymptotes. And one more thing to note since we know that the hyperbola opens up to the left and to the right, we can go ahead and draw the shape of the hyperbola with respect to the ascent totes. And the hyperbola is going to have a rough shape like this. OK. Now we have the general form or the general shape of the equation of the hyperbola with the center and vertices located. The next thing to do is to find the locations of the. Now the are going to be some distance C to the right and to the left of the center of the hyperbola. And we have an equation for hyperbolas that help us find this distance the equation that we have is C squared is equal to A squared plus B squared. We already know the values of A squared and B squared A squared is 36 B squared is nine. So if we plug this into the equation, we get C squared is equal to 36 plus nine, 36 plus nine is going to give us the value of 45. So C squared is equal to 45. And if we take the square root of both sides of the equation, we get C is equal to three times the square root of five. What this means is that the are gonna be located a distance of three square to five to the right and to the left of the center of the hyperbola. And that means we're going to be adding them subtracting the value of three square to five to the X value of the center of the hyperbola. That means the locations of the are going to be one plus three square root of five comma zero and one minus three square root of five comma zero. These are gonna be the locations for the. Now, what we need to do is we need to find the equations of the vert diagonal asymptotes. So recall that we said that the location of the hyperbola is now located at the origin and the major axis is the X axis. That means the general form of the equations of the diagonal asymptotes is going to be Y is equal to plus or minus B over A times the quantity X minus H plus K, we know the values of B and A B is equal to three and A is equal to six. Furthermore H is equal to one and K is equal to zero. So if we plug this into the general form of the equations of the diagonal ascent totes, we end up getting two equations. Y is equal to positive 3/6 times the quantity X minus H which is X minus one plus K, which is plus zero. And we also get Y is equal to negative 3/6 times the quantity X minus one plus zero. Now 3/6 is going to simplify to give us one half and negative 3/6 is going to simplify to give us negative one half. On the left hand side, if we distribute a half to the quantity X minus one, we get Y is equal to one half, X minus one half. And since we're adding zero at the end, we don't have to include the zero. So this is going to be the first equation for the diagonal Asymptote. And on the right hand side, if we distribute negative one half to X minus one, we get Y is equal to negative 1/2 X plus 1/2. And again, since we're adding zero, we don't have to include that value. So this is going to be the second equation for the diagonal Asymptote. So just to recap, we found the equations of the diagnose totes, we found the locations for the and we have the rough sketch of the hyperbola with the vertices and the center graft. And 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.

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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)^2/25−y^2/16=1

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