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Math Help - Ellipses word problem

  1. #1
    Junior Member Greymalkin's Avatar
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    Ellipses word problem

    Ellipses word problem-trig1.png

    Problem above, from what I understand you are to find the vertical distance to an edge that is 6 feet from a vertex.
    I'm trying to triangulate from the focus distances to the point to create an equation I can equate.
    Using distance from the focus(c= \sqrt{(25^2-14^2)}=20.71
    Setting the center at the origin, the distance from 2 foci = 2(\sqrt{20.71^2+14^2}=50
    Therefore the point 6 feet from the vertex, which is 19 ft from the origin, and is 39.71ft from the last focus, is equated by: 50-(39.71^2+b^2=c^2)
    This is where my approach halts, is my approach wrong? Or is there something I am just not seeing?

    Edit answer is ~9.1
    Last edited by Greymalkin; August 21st 2012 at 02:24 PM.
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  2. #2
    GJA
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    Re: Ellipses word problem

    Hi, Greymalkin.

    I think your approach is good, nice work! I think a small hint will get things going in the right direction.

    Since you're taking the origin to sit "in the middle" so-to-speak, we know that x and y must satisfy the equation

    \frac{x^{2}}{(25)^{2}}+\frac{y^{2}}{(14)^{2}}=1,

    where 25 is the major radius of the ellipse, and 14 is the minor radius. What you determined in your work was that the x coordinate of the point we're interested is x=-19. Our goal is to know what the height y is at this point; to determine this we can use the ellipse equation above since we know what  x is at the point of interest.

    Does this help? Let me know if anything is unclear.

    Good luck!

    Edit: I was thinking of the bank on the left of the picture, that's why I have x=-19 above, but if you wanted to consider the right bank you could use x=19. It's all academic because we're going to square things in the equation of the ellipse anyways!
    Last edited by GJA; August 21st 2012 at 05:30 PM.
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