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OpenStudy (anonymous):

Which set of parametric equations represents the following (Attached) Possible Answers: A. x = -1 + √15 secθ y = 6 + 7tanθ B. x = 1 + √15 secθ y = -6 + 7tanθ C. x = -6 + √15 secθ y = 1 + 7tanθ D. x = 1 + 7secθ y = -6 + √15 tanθ E. x = -6 + 7secθ y = 1 + √15 tanθ

OpenStudy (anonymous):

jimthompson5910 (jim_thompson5910):

The center is at the midpoint of either the foci or the vertices (along the transverse axis) So let's find the midpoint between the two vertices X coordinate of the midpoint = (x1+x2)/2 = (-6+8)/2 = 2/2 = 1 Y coordinate of the midpoint = (y1+y2)/2 = (-6+(-6))/2 = -12/2 = -6 So the midpoint is (1,-6) Which makes the center (1,-6). Since (h,k) is the center in general, we can say (h,k) = (1,-6) and h = 1, k = -6 ---------------------------------------------------------------- Now let's find the length of the major axis This is the distance between the two vertices. The two vertices are (-6,8) and (8,-6). Because the y coordinates are equal, the distance is simply 8 - (-6) = 14 units. So the major axis is 14 units. So 2a = 14 ----> a = 7 Note: the value of 'a' is half of the major axis (or a = length of semi-major axis) ----------------------------------------------------------------- Now find the distance from the center to one focus. This distance is 9 - 1 = 8 units. So c = 8 Now use a = 7 and c = 8 to find b in the equation below a^2+b^2 = c^2 7^2 + b^2 = 8^2 49+b^2 = 64 b^2 = 64-49 b^2 = 15 b = sqrt(15) -------------------------------------------------------- Now in general, the parametric form of any hyperbola is x = h+a*sec(theta) y = k+b*tan(theta) In this case, h = 1, k = -6, a = 7 and b = sqrt(15) So the parametric form of this hyperbola is \[\Large x = 1 + 7\sec(\theta)\] \[\Large y = -6 + \sqrt{15}\tan(\theta)\] So the answer is choice D

OpenStudy (anonymous):

Thank you so much! I really appreciate the help!

jimthompson5910 (jim_thompson5910):

you're welcome :)

OpenStudy (anonymous):

Then, in general, what would be the parametric form of a line or conic?

jimthompson5910 (jim_thompson5910):

parametric form of a line: x = a+b*t y = c+d*t parametric form of a conic: depends on the conic (it will change as the conic type changes)

OpenStudy (anonymous):

Okay. Then how would I find d and t? Would I find a and c the same way as I did the other one? Here the question I'm working on:

OpenStudy (anonymous):

From what you did before, i figured out that h = -7 and k = -2, but I'm not quite sure how what we did before is applicable to the rest of the question.

jimthompson5910 (jim_thompson5910):

Good, you got the center correct. Now you just need to find the length of the major and minor axes. Do you know how to do this?

OpenStudy (anonymous):

From what you did in the first one, I think that the major axis would by 18. I'm not sure about the minor.

OpenStudy (anonymous):

be*

jimthompson5910 (jim_thompson5910):

Sorry got distracted for a bit. You nailed it. So 2a = 18 ---> a = 9 Note: In this case, the major axis is along the horizontal. This is why 'a' is the length of the semi-major axis. ------------------------------------------------------- Now we must find the distance from one focus to the center. One focus is (-14,-2) and the center is (-7,-2). Therefore, the distance is simply |-14 - (-7)| = |-14 + 7| = |-7| = 7 So the distance from the center to either focus is 7 units, which means that c = 7 Now use the idea that b^2 = a^2 - c^2 to solve for b b^2 = a^2 - c^2 b^2 = 9^2 - 7^2 b^2 = 81 - 49 b^2 = 32 b = sqrt(32) b = 4*sqrt(2) Therefore, the answer is \[\Large x = -7 + 9\cos(\theta)\] \[\Large y = -2 + 4\sqrt{2}\sin(\theta)\]

OpenStudy (anonymous):

It does. I didn't mind waiting. Thank you so much! So, would I do the same sort of thing for this type of problem? Which set of parametric equations represents the graph of the following rectangular equation using t = 1 - x. y = x2 + 2

jimthompson5910 (jim_thompson5910):

are you given a graph?

OpenStudy (anonymous):

No

jimthompson5910 (jim_thompson5910):

do you have answer choices?

OpenStudy (anonymous):

Yes... give me a minute

jimthompson5910 (jim_thompson5910):

alright

OpenStudy (anonymous):

A. x = t + 1 y = (t + 1)^2 - 2 B. x = t - 1 y = (t - 1)^2 - 2 C. x = t - 1 y = (t-1)^2 + 2 D. x = 1 + t y = (1 + t)^2 + 2 E. x = 1 - t y = (1 - t)^2 + 2

jimthompson5910 (jim_thompson5910):

oh ok, I see what they want now. Thanks

jimthompson5910 (jim_thompson5910):

We're given t = 1-x. Or they're telling us to use this given fact. Solve for x: t = 1 - x t - 1 = -x -x = t - 1 x = -t + 1 x = 1 - t --------------------------- Now move onto the actual equation y = x^2 + 2 y = x^2 + 2 y = (1 - t)^2 + 2 ... Replace 'x' with 1 - t (since x = 1-t ) So the set of parametric equations is x = 1 - t y = (1 - t)^2 + 2 which is choice E

OpenStudy (anonymous):

Thank you soooo much. I really appreciate that you took the time to help me with these! Have a great night!!

jimthompson5910 (jim_thompson5910):

you too, I'm glad I could help you out

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