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  • one year ago

Johanna attempted to base her hyperbola off of maurice's selected y-axis vertices. Since the foci are the same, that must be the greatest distance from the center. Otherwise, the major axis will change and therefore so will the focal points. Johanna has decided to choose y-axis vertices of (0, 2) and (-2, 0).

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  1. anonymous
    • one year ago
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    This is the previous half of the 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. Maurice wants to use elliptical flower beds, so here is the equation: He has to use the equation --> a^2-b^2=c^2. C= 3 and I made a equal to 6. 6^2 - b^2 = 3^2. 36-b^2 = 9. b^2 = 27 b=5.196 --> x^2/a^2 + y^2/b^2 --> x^2/36 + y^2/27 = 1 Johanna wants to use hyperbolic flower bed, this is her equation: A will = 2 2^2+b^2=3^2 --> 4 + b^2 = 9 --> b^2 = 5 --> b = 2.236 x^2/a^2-y^2/b^2 = 1 --> x^2/2^2 - y^2 / 2.236^2 = 1

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