Question

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.

Answer 1

medal and fan!!!

Answer 2

@IMStuck hey can you help

Answer 3

@zepdrix hi can you help

Answer 4

\((3,0)\) and \((–3, 0)\) are to be the focal points of the ellipse right?

Answer 5

for this you want
\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\] 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 \((3,0)\)

Answer 6

check the nice picture here to see that it works
http://www.wolframalpha.com/input/?i=ellipse+x^2%2F25%2By^2%2F16%3D1

Answer 7

for the hyperbola it will look like
\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\] and you want \(a^2+b^2=3^3\) simplest way i can think to do it is to make \(a^2=8,b^2=1\) and use
\[\frac{x^2}{8}-y^2=1\] but you have other choices

Answer 8

Perfect!! thank you @satellite73

Answer 9

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

Answer 10

yw

Answer 11

why is the formula a^2+b^2=3^3 ? @satellite73

Answer 12

why is a very good question
it will take a lot of explaining however
a better explanation is here
http://www.purplemath.com/modules/hyperbola2.htm

Answer 13

well not why... how

Answer 14

can you please explain the hyperbola part

Answer 15

@hero can you explain the hyperbola part please