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. (y+2)^2/4−(x−1)^2/16=1

Accepted Answer

Hello, today we're gonna be graphing the equation of the hyperbola. Then we're gonna be finding the equations for the Diago asymptotes and the coordinates for the. So what we are given is Y minus four squared over minus X plus three squared over 100 is equal to one. So the equation that's given to us is already in the standard form of the equation of the hyperbola. So there's a few things that we can make note of from this equation. The first thing is that the Y and X quantity are given to us as binomials. What that means is that the equation of the hyperbola is not located at the origin and the center of the hyperbola is going to be defined as H comma K. Furthermore, the Y term is the leading term on the left hand side of the equation. So that means that the major axis for the equation of the hyperbola is going to be the Y axis. So since we know that this is not located at the origin and the major axis of the Y axis, the general form of the equation of the hyperbola is going to be Y minus K squared over A squared minus X minus H squared over B squared is equal to one. And this is where we have a hyperbola that opens up to the top and to the bottom. The distance of the vertices from the center of the hyperbola are some distance A and the width of the as in total box is some distance B away from the center. So let's go ahead and use the general form of the equation of the hyperbola to find our AM B values and to find the center of the hyperbola. So in the a square spot, we have the value of 64. And in the B square spot, we have the value of 100. So we can go ahead and set A squared equal to B squared equal to 100. If we take the square roots of both of the equations, we get A is equal to plus or minus eight and B is equal to plus or minus 10. Now let's go ahead and find the center of the hyperbola. So in order to get the H value, we're gonna take a look at the X quantity given to us, we are given X plus three squared. This is not in the general form of X minus H squared, but it can be rewritten as X minus negative three squared. This is in the general form of X minus H squared and this is where H is equal to negative three. Now let's get the K value, the Y value is given to us as Y minus four squared. This is in the general form of Y minus K squared. And this is where K is equal to positive four. And that means that the center of the hyperbola is given to us as negative three, comma positive four. So now that we have the center and the AM B values, we can use this information to create a rough sketch of the hyperbola. So we know that the center of the hyperbola is located at negative three comma four. The vertices are of the hyperbola are gonna be located eight units above and below the center. And that means that the vertices are gonna be located at negative three comma 12 and negative three comma negative four. And the width of the as in total box is gonna be defined by units to the right and to the left of the center. So that means that the width is gonna be located at seven comma four and negative comma four. We can go ahead and connect these four dots together to create our M total box. And we can use this box to create a rough sketch of the diagonal asymptotes. Now we know that this hyperbola opens up to the top and to the bottom. So we can go ahead and draw the shape of the hyperbola with respect to the ascent totes and the hyperbola is gonna have a rough shape just like this. All right. So now we have the general shape of the hyperbola. The next thing we need to do is we need to find the equate the locations of the. Now, the foci are located some distance C above and below the center of the hyperbola. We need to go ahead and find the distance of C and we have an equation for a hyperbola that allows us to do this. For the, we have the equation C squared is equal to A squared plus B squared. We know that the values of A squared and B squared are going to be 64 100. So if we substitute this into the equation, we get C squared is equal to 64 plus 64 plus 100 is going to give us 1 64. So C squared is equal to 1 64. And if we take the square root of both sides of the equation, we get C is equal to two times the square root of 41. So what this means is that the site are located a distance of two square root 41 above and below the center of the hyperbola. And what this means is that we're going to be adding and attracting the value of two square root 41 to the Y value of the center of the hyperbola. That means the locations of the are going to be negative three comma four plus two square root of 41 and negative three comma four minus two square root of 41. 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 asymptotes. Now, one thing to note from earlier is that the center is not located at the origin and the major axis is the Y axis. With that being said, the general form of the equations of the diagonal asy totes are going to be Y is equal to plus or minus A over B times the quantity X minus H plus K. Now we know what our AM B values are going to be A is eight and B is 10, H is equal to negative three and K is equal to four. So if we plug this into the general form of the equation of Diago ascent totes, we end up getting two equations. Y is equal to positive 8/ times the quantity X minus negative three, which is X plus three plus K which is plus four. And we get the negative version which is Y is equal to negative 8/10 times the quantity X plus three plus four. Now 8/10 is going to reduce to give us 4/5 and negative 8/10 is going to reduce to give us negative for over five. So if we distribute 4/5 into the quantity X plus three, we get Y is equal to 4/5, X plus 4/5 times three, which is going to give us 12/5 plus four. And on the right hand side, if we distribute negative 4/5 to X plus three, we get Y is equal to negative 4/5, X minus 4/5 times three, which is going to give us 12/5 plus four. On the left hand side, if we add 12/5 with four, we get the value of 32/5. So the equation of the first diagonal S tote is going to be 4/ X plus 32/5. And if we added four to negative 12/5, we get the value of 8/5. So the second diagonal asym coat asym tote is going to be defined as negative 4/5 X plus 8/5. So just to summarize, we just found the equations for the diagonal ascent totes, we also found the coordinates for the sign. And up above, we created the rough shape of the equation of the hyperbola with the center and the vertices labeled. 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. (y+2)^2/4−(x−1)^2/16=1

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