anonymous
  • anonymous
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.
Geometry
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
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anonymous
  • anonymous
@Hero plz help
anonymous
  • anonymous
i can help you :)
anonymous
  • anonymous
OMG THANK YOU SO MUCH I NEED SLEEP OMG

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More answers

anonymous
  • anonymous
lol ok :)
anonymous
  • anonymous
(3,0) and (–3,0) are to be the focal points of the ellipse right?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
x^2/25+y^2/16=1 is that the equation for the elliptical flower bed?
anonymous
  • anonymous
for this you want \[\frac{ x2 ^{} }{ a2 } + \frac{ y2 }{ b2 } = 1\] and you want \[a^{2} - b^{2} = 3^{2}\] the easiest way to do that is to use the famous 3−4−5 right triangle and make a=5,b=4 so c=3 and use \[\frac{ x^{2} }{ 5^{2} } + \frac{ y^{2} }{ 4^{2} } = 1\] that will make your foci (−3,0) and (3,0)
anonymous
  • anonymous
ok I got that part but what about the hyperbolic flowerbed
anonymous
  • anonymous
can you try
anonymous
  • anonymous
I can't make up the value of A
anonymous
  • anonymous
like I know x^2/a^2-y^2/b^2=1 and a^2+b^2=3^2
anonymous
  • anonymous
for the hyperbola it will look like x2/ a2−y2/b2=1 and you want a2+b2=33 simplest way i can think to do it is to make a2=8,b2=1 and use x2/8−y2=1 but you have other choices
anonymous
  • anonymous
how did you get 33?
anonymous
  • anonymous
no its \[3^{3}\]
anonymous
  • anonymous
ok
anonymous
  • anonymous
medal and fan please
anonymous
  • anonymous
yeah but one more question how do you graph the second one?
anonymous
  • anonymous
here is a nice graph with both together if you need one http://www.wolframalpha.com/input/?i=+x^2%2F8-y^2%3D1%2Cx^2%2F25%2By^2%2F16%3D1
anonymous
  • anonymous
ok thank you so much!!
anonymous
  • anonymous
your welcome
anonymous
  • anonymous
medal and fan?
anonymous
  • anonymous
yeah i already did
anonymous
  • anonymous
ok thanks :)

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