yeval76
  • yeval76
***WILL FAN AND MEDAL*** There are two fruit trees located at (3,0) and (−3, 0) in the backyard plan. Maurice wants to use these two fruit trees as the focal points for an elliptical flowerbed. Johanna wants to use these two fruit trees as the focal points for some hyperbolic flowerbeds. Create the location of two vertices on the y-axis. Show your work creating the equations for both the horizontal ellipse and the horizontal hyperbola. Include the graph of both equations and the focal points on the same coordinate plane. @ChiefArnav
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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yeval76
  • yeval76
@nincompoop
yeval76
  • yeval76
@Abhisar please help
yeval76
  • yeval76
@Loser66

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Loser66
  • Loser66
|dw:1439645146687:dw|
Loser66
  • Loser66
You are "Creating" the ellipse. Hence it is up to you to pick the a and b such that \(c^2= a^2-b^2\). Your c is 3, hence \(c^2 =9\) you can pick a, b satisfy that equation. Which pair do you want to pick?
Loser66
  • Loser66
@Penguin7 This problem is horizontal, your problem is vertical. I work for him first, you convert everything to vertical. That is it.
Loser66
  • Loser66
@yeval76 If you don't help yourself, you don't work for your own problem. You are better move to @Penguin7 post and convert to horizontal case from her case. I prioritize for the students who join me on their stuff.
yeval76
  • yeval76
Im so sorry I didn't notice that you had responded. I will join in and work for my problem... Please help me @Loser66
yeval76
  • yeval76
I pick that a=6
Loser66
  • Loser66
Ok, answer me, pick a , b such that 9 = a^2 -b^2
yeval76
  • yeval76
9 = 6^2 -b^2
yeval76
  • yeval76
9 = 36- b^2
Loser66
  • Loser66
a =6 , hence a^2 =36, so b =?
yeval76
  • yeval76
b^2 = 27 but b = 5.196
Loser66
  • Loser66
ok, how about a =5 and b=4? do they work?
Loser66
  • Loser66
Your pick is PERFECTLY ok. I just suggest the easiest numbers to work with.
yeval76
  • yeval76
If a = 5 and b = 4, it would be the following: 25-16=9
Loser66
  • Loser66
Yes, hence your ellipse equation will be??? just plug into the standard form for HORIZONTAL one.
Loser66
  • Loser66
http://www.math.binghamton.edu/grads/kaminski/math222_spring12/Sect10.5_ShiftedConics.pdf
Loser66
  • Loser66
Print it out to use later. It is helpful
Loser66
  • Loser66
Look at Horizontal Ellipse row. the formula right on the 3rd column.
Loser66
  • Loser66
The net is so bad. I assume you cannot access. but you know how to do this part, right? Now, move to hyperbola.!!!
Loser66
  • Loser66
The same token, but for hyperbola, \(c^2 = a^2 \color{red}{+} b^2\) You can create a and b such that \(9= a^2 +b^2\)
Loser66
  • Loser66
On this case, no way for you to choose both a, b are integers. If you choose a = 1, then \(b= 2\sqrt2\), If you choose a=2, then \(b=\sqrt5\) Whatsoever, you pick 1 pair, a and b, then plug back to row HORIZONTAL HYPERBOLA on the paper to get the equation. Good luck.
Loser66
  • Loser66
|dw:1439646772690:dw|

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