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Finding equation of ellipse

  1. Jan 30, 2005 #1
    1. Find an equation of an ellipse with vertices at points (-2, 5) and (-2, -3), containing point (0, 3).

    2. A bridge is to be constructed across a river that is 200 feet wide. The arch of the bridge is to be semielliptical and must be constructed so that a ship less than 50 feet wide and 30 feet high can pass safely through the arch.
    (a) Find an equation in standard form of an ellipse that encompasses the semielliptical arch of the bridge.
    (b) Determine the exact height of the semielliptical arch in the middle of the bridge.


    I have a feeling these aren't as complicated as I'm trying to make them...usually arent. I hate ellipses. Maybe if I understood them better it would help... Thanks--Jennifer
     
  2. jcsd
  3. Jan 31, 2005 #2
    Do you have a textbook?
     
  4. Jan 31, 2005 #3
    Yes, I have a textbook but my teacher is not going by the book. This stuff isn't really in there.
     
  5. Jan 31, 2005 #4

    HallsofIvy

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    The formula for an ellipse (with axes parallel to the x and y axes and I doubt that you are doing the much more difficult case where they are not!) is
    [tex]\frac{(x-x_0)^2}{a^2}+ /frac{(y-y_0)^2}{b^2}= 1[/tex]
    where (x0,y0) is the center of the ellipse and a and be are the lengths of the "semi-axes" in the x and y directions. Of course, you are not GIVEN any of those values so you will have to try to calculate them.
    Two vertices are (-2,5) and (-2, -3) so (assuming the axes are parallel to the x and y axes!) they are on one axis and the center of the ellipse is at the center of that interval: (-2, (5-3)/2)= (-2, 1). 5-1= 4 and 1-(-3)= 4 so "b" the length of the semi-axis parallel to the y-axis is 4. You now know x0= -2, y0= 1, and b= 4. You still don't know a. You have the equation
    [tex]\frac{(x+2)^2}{a^2}+ \frac{(y-1)^2}{16}= 1[/tex].

    You also are told that (0, 3) is on the ellipse. Putting x= 0, y= 3 into your equation leaves an equation with the single "unknown", a. Solve for a (actually, you only need to find a2) and put that into the equation.

    2. Choose a coordinate system so (0,0) is at the middle of the river and the x-axis runs across the river (y-axis is up of course). That has the advantage of making the center of the ellipse at (0,0). Obviously the length of the ellipse is the width of river, 200 feet. The length of the "semi-axis" is half that: 100 feet. With that information, you know that the equation of the ellipse is
    [tex] \frac{x^2}{1000}+ \frac{y^2}{b^2}= 1[/tex]
    and you only have to find b.
    You are told "a ship less than 50 feet wide and 30 feet high can pass safely". Assuming the ship goes right down the center of the river, a ship 50 feet wide would extend 25 feet on either side. I THINK the problem intends that the "30 feet high" might be any where on the boat so you would need to have the ellipse at least 30 feet high 25 feet from the center: that is (25, 30) must be on the ellipse.
    Put x= 25, y= 30 into your formula and solve for b (which will also answer the second part of the question).
     
  6. Feb 1, 2005 #5
    Thanks BuT! Ok--How do I go about finding the equation of only the top part of the ellipse-the positive y part. I understand how to come at the equation for the entire thing but I just need the top. I figure it has something to do with solving for y but I don't know what to do with that even if I do.
     
  7. Feb 1, 2005 #6

    HallsofIvy

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    If you cannot solve a quadratic equation then you should not be taking precalculus!

    Your teacher will quite properly assume that you can do basic algebra.
     
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