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

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

Thank you so much! I really appreciate the help!

you're welcome :)

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

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)

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:

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.

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?

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

be*

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 \[\Large x = -7 + 9\cos(\theta)\] \[\Large y = -2 + 4\sqrt{2}\sin(\theta)\]

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

are you given a graph?

No

do you have answer choices?

Yes... give me a minute

alright

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

oh ok, I see what they want now. Thanks

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

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

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

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