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−1)^2−(y−2)^2=3

Accepted Answer

Hello, today we're gonna be graphing the equation of the hyperbola. Then finding the equations for the diagnose totes and the coordinates for the. So what we are given is X plus three squared minus Y plus five squared is equal to five. Now, the first thing we want to do in simplifying this equation is you want to set this equation equal to one in order to do that, we're gonna divide every term in the equation by five. Doing so allows us to rewrite the equation of the hyperbola as X plus three squared over five minus Y plus five squared over five is equal to one. Now, there's a few things to note about this equation of the hyperbole. The X and Y values are given to us as binomials. And what that means is that the center is not located at the origin, the center is going to be defined as H comma K. Another thing to note is that the X term is the leading term in this equation of the hyperbola. And what that means is that the major axis for the hyperbola is going to be the X axis. So what this means is that we have a hyperbole with respect to the X axis that opens up to the left and to the right, the vertices are defined with an A variable and the height of the as in total box is defined using A B variable and A and B A and B represent the distances from the center of the hyperbola. So another thing to note is that since this is a hyperbole with the center, not at the origin and major axis, the X axis, the general form of the equation of the hyperbole is going to be X minus H squared over A squared minus Y minus K squared over B squared is equal to one. So let's go ahead and use the general form of the equation of the hyperbola in order to get our A and B values and the center of the hyperbola, one thing to note is that five takes up the spot of A squared and five takes up the spot of B squared. What that means is that A squared is equal to five and B squared is equal to five. And if we take the square root of both sides of both of the equations, we get A is equal to plus or minus the square root of five and B is equal to plus or minus the square root of five. So we have our distances for AM B. Now we need to find the center of the hyperbola. So in order to get the center, let's take a look at the X value. The X value is given to us as X plus three squared. This is currently not in the general form of X minus H squared, but it can be rewritten as X minus negative three quantity squared. This is in the general form of X minus H squared where H is equal to negative three. Now let's take a look at the Y value in order to get a K value, we are given Y plus five squared. This is not in the general form of Y minus K squared, but it can be rewritten as Y minus negative five squared. This is in the general form of Y minus K squared where K is equal to negative five. So now that we have H is equal to negative three and K is equal to negative five, that means the center of the hyperbola is going to be located at negative three comma negative five. So now that we have the center of the hyperbola and we have the distances for the vertices and the height of the as in total box, let's go ahead and create a rough sketch of our, of our hyperbola. So we know that the center is located at negative three comma negative five. The vertices are gonna be located a distance of square to five to the right and to the left of the center. So that means that the vertices to the right is gonna be located at negative three plus the square root of five comma negative five. And the other vertices is gonna be located to the left at negative three minus the square root of five comma negative five. And the height of the as in total box is gonna be square root five units above and below the center of the hyperbola. So it's going to be located at negative three comma negative five plus the square root of five and at negative three comma negative five minus the square root of five. So now what we can go ahead and do is connect the four dots and create our s in total box. We can even use this box to create a rough sketch of the diagonal asymptotes. OK. Now that we know that the uh that the hyperbola is going to equal or open up to the right and to the left, that means that we can use the ace totes to draw the rough shape of the hyperbola. And that means the hyperbola is gonna roughly look like this. All right. So now that we have the rough shape of our hyperbola, there's two more things that we need to get, we need to get the location of the foci which in this case are gonna be located some distance to the right and left of the vertices. And we need to get the equations for the diagonal asymptotes. Let's go ahead and start by looking for the locations of the. Now the are some distance C from the center of the hyperbola. We have an equation that helps us find this distance of C that is C squared is equal to A squared plus B squared for hyperbolas. And we know what A squared and B squared are equal to both of these values are equal to five. So if we plug this into the equation, we get C squared is equal to five plus five, five plus five is going to give us 10. So C squared is equal to 10. And if we take the square roots of both sides of the equation, we get C is equal to plus or minus 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 hyperbole. And what that means is that we're going to be adding and subtracting the value of the square root of 10 from the X value of the center of the hyperbola. That means the locations of the are going to be negative three plus the square root of 10, comma negative five and negative three minus the square root of 10 comma negative five. So these are gonna be the coordinates for the sign. Now, what we need to do is we need to get the equations for the diagonal Asy totes. Now two things to note this hyperbola is not located at the origin and the major axis is the X axis. So the general form of the equations of the diagonal ascent totes is going to be Y is equal to plus or minus B over A times the quantity X minus H plus K. We already know what our B N A values are B and A are both square root of five. We know that H is negative three and we know that K is negative five. So if we plug this in to the general form of the equation of the Asy Tode, we get two different equations. We get Y is equal to positive square root of five over squared of five times the quantity X minus negative three, which is going to give us plus three minus five. And we also get Y is equal to negative square root five over square root five times the quantity X plus three minus five. So there's a bit of simplification that we can do first square root five over square root of five is gonna to simplify to give us one and negative square root five over square root five is going to simplify to give us negative one. So what we are left with is Y is equal to the quantity X plus three minus five and Y is equal to negative X plus three minus five. On the left hand side, three minus five. Is going to give us negative two. So we get Y is equal to X minus two. This is going to be the first equation for one of the diagonal asy totes. For the other equation, we're going to distribute negative one into X plus three and we get Y is equal to negative X minus three minus five. Finally negative three minus five is going to give us negative eight. So we get Y is equal to negative X minus eight. And this is going to be the second equation for the diagonal scent tote. So just to summarize, we just calculated the equations for the diagonal asymptotes, we have the locations of the and we created a rough sketch of the hyperbola. So with that being said, these four pieces of information are going to be our solution. 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−1)^2−(y−2)^2=3

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