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anonymous
 one year ago
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 yaxis. 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.
anonymous
 one year ago
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 yaxis. 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.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OMG THANK YOU SO MUCH I NEED SLEEP OMG

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(3,0) and (–3,0) are to be the focal points of the ellipse right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x^2/25+y^2/16=1 is that the equation for the elliptical flower bed?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for 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
 one year ago
Best ResponseYou've already chosen the best response.0ok I got that part but what about the hyperbolic flowerbed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can't make up the value of A

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like I know x^2/a^2y^2/b^2=1 and a^2+b^2=3^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for 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
 one year ago
Best ResponseYou've already chosen the best response.0medal and fan please

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah but one more question how do you graph the second one?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0here is a nice graph with both together if you need one http://www.wolframalpha.com/input/?i=+x^2%2F8y^2%3D1%2Cx^2%2F25%2By^2%2F16%3D1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok thank you so much!!
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