Question

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

Answer 1

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

Answer 2

Thank you so much! I really appreciate the help!

Answer 3

you're welcome :)

Answer 4

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

Answer 5

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)

Answer 6

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:

Answer 7

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.

Answer 8

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?

Answer 9

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

Answer 10

be*

Answer 11

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

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

Answer 12

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

Answer 13

are you given a graph?

Answer 14

No

Answer 15

do you have answer choices?

Answer 16

Yes... give me a minute

Answer 17

alright

Answer 18

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

Answer 19

oh ok, I see what they want now. Thanks

Answer 20

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

Answer 21

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

Answer 22

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