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.

Exercises 53-56 give equations for hyperbolas and tell how many units up or down and to the right or left each hyperbola is to be shifted. Find an equation for the new hyperbola, and find the new center, foci, vertices, and asymptotes.

x²/4 − y²/5 = 1, right 2, up 2

Accepted Answer

Given the hyperbola, X2 divided by 16 minus Y2 equals 1, shift it right 5 units and down 4 units. Determine the equation of the new hyperbole and find its center, vertices, foci in equations for the asymptotes. Now, to solve this, let's first shift our equation around. So we're shifting it right 5 units, which means we'll first replace X with X minus 5, and we'll just keep the squared and the 16 in the denominator. We'll do the same for y. Set the shifting down 4 units. We will put Y + 4. Squared And there's no number underneath, so we'll just leave it. This equals one. Now, because we shifted this, we have a new center. Before the center was at 00. Now that we've shifted it, our new sinner is given by -5, or rather 5 because of the shift to the right. And -4. And so now we have our sinner. Which is given by HK. Now let's also note that our A, in this case. Will be found with A squared, which is 16. This is underneath the X. Our B2 is 1, which is underneath the Y. There's not a number there, but since there's not a number, it's just 1. This means A equals 4 and B equals 1 when you take the square root of both. We can now use this to find C, which we'll use to find the fosite. We have C2 equals A2 + B2, which means we have 16 + 1 or 17. C is then the square root of 17, and that does not simplify. Now we have our center, we have A, B, and C. We can use these to find our vertices and foesite. Now, this is a horizontal hyperbole because A or the larger value is underneath the X. So, for the vertices. Of a horizontal hyperbola. We will take H and add plus or minus A, and keep the value for K in the points. That means we have 5 plus or minus 4, -4. This gives us two different points, being 9-4. And 1-4. Now we can use this to find the foresite. The fosci very similarly to the vertices, we will take. H Plus or minus c In K and the point. We have 5 plus or minus the square root of 17-4. Which gives us 5 + 2 of 17-4. And 5 minus the square of 17, -4, and those don't simplify. Finally, we can find our asymptotes. The asymptotes for a horizontal hyperbola are given by Y minus K. Equals plus or minus B divided by A X minus H. We can then write why. Plus 4. Equals plus or minus 1/4 x minus 5. We can then find, solve for Y and simplify it. We ended up getting Y equals 1/4. X. Plus 5/4. And Y equals -14 X minus 11/4. We now have everything for the answer to our problem. We have our equation. Our center. Our asymptotes. Our vertices and finally our fosite. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

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