Shifting Conic Sections

You may wish to revie...

Question

Shifting Conic Sections

You may wish to review Section 1.2 before solving Exercises 39-56.

The hyperbola (y²/4) − (x²/5) = 1 is shifted 2 units down to generate the hyperbola (y + 2)²/4 − x²/5 = 1.

a. Find the center, foci, vertices, and asymptotes of the new hyperbola.

Accepted Answer

Hello. In this video, the hyperbola x2 divided by 9 minus y2 divided by 16 equal to 1 is shifted 3 units to the right to generate the equation of the hyperbolax minus 3 quantity squared divided by 9 minus y2 divided by 16 is equal to 1. We want to determine the fosci, center, vertices, and asymptotes of the new hyperbola. So let's go ahead and write down the equation of the transformed hyperbola. We have X minus 32. Divided by 9 minus y2 divided by 16 is equal to 1. Now, if we were to compare this to the general form of the hyperbola, the general equation of the hyperbola is going to be X minus H quantity 2 divided by A2 minus Y minus K2 divided by B2 is equal to 1. Now this matches the equation of a hyperbola where the x axis is the major axis. So this tells us that the shape of the hyperbola is going to open to the left and to the right. But more importantly, because we're shifting the hyperbola 3 units to the right, this means that this is a hyperbola that is not centered at the origin. So the center is going to be defined. As HK. Now if we were to compare. Our general hyperbola to the transformed one in the problem, we see that x minus 3 matches the quantity of X minus H, and this tells us that the value of h is going to equal to 3. Now, Y squad is in the form of Y minus K, but this is when K is equal to 0. So this tells us that the center of the hyperbola is going to be located at 3.0, and that is going to be the first solution to this problem. But now what we want to do is we want to go ahead and define the vertices. So let's find the vertices of the hyperbola next. Now the vertices are going to be defined as h plus or minus A, 0, because this hyperbole lies on the x axis. Now, here, we know that A2 is equal to 9 and B2 is equal to 16. If we take the square root of both of the values, we get A is equal to 3 and B is equal to 4. And we also know that the value of H is going to be 3. So by plugging this into the identity of the vertices, we can redefine the vertices to be 3 plus or minus 3.0, and that is going to give us 2 points for the vertices. The vertices are going to be located at the origin and at 6.0. This is going to be the 2nd solution to this problem. Now we're going to identify the foci. Now the foci in this case are going to be defined as h plus or minus c, 0, and c is defined by the property c2 is equal to a2 + b2. Now we know A2 is equal to 9 and B2 is equal to 16. So by plugging in these values, we get C2 is equal to 9 + 16. That is going to simplify to be c squared is equal to 25, and by taking the square root of both sides of this equation we get c is equal to 5. Again, h is equal to 3. So by plugging in these values, the foci is going to be defined as 3 plus or minus 5.0, and that is going to give us two locations for the foci. We are going to have one foci be located at -2.0, and the second one will be located at 8.0, and this is going to be a third solution to the problem. Now the final thing we need to identify is the equation of the asymptotes. Now the asymptotes in this case are going to be defined as Y is equal to positive and negative, B divided by a multiplied by X minus H. Now we know that B is equal to 4 and A is equal to 3. So this is going to define our asymptotes as Y is equal to positive and negative, 4 divided by 3, multiplied by x minus H, which will be x minus 3. So, this is the simplest form of the asymptodes, and that means we are going to have two equations. We are going to have y is equal to 4/3 multiplied by x minus 3. And we will have y is equal to -4/3 multiplied by x minus 3, and these are going to be the final solutions to the given problem. So just to recap. Our center is going to be located at 30. We have vertices located at the origin and 60. Foci are located at -20 and 80, and the asymptotes are going to be positive and 4/3 multiplied by x minus 3. With that being said, the overall solution to this problem is going to be option A. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

Shifting Conic SectionsYou may wish to review Section 1.2 before solving Exercises 39-56.The hyperbola (y²/4) − (x²/5) = 1 is shifted 2 units down to generate the hyperbola (y + 2)²/4 − x²/5 = 1.a. Find the center, foci, vertices, and asymptotes of the new hyperbola.

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Shifting Conic Sections

You may wish to review Section 1.2 before solving Exercises 39-56.

The hyperbola (y²/4) − (x²/5) = 1 is shifted 2 units down to generate the hyperbola (y + 2)²/4 − x²/5 = 1.
a. Find the center, foci, vertices, and asymptotes of the new hyperbola.