Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Ellipse Circumscribed In Triangle

  1. Dec 12, 2007 #1
    Let P(x,a) and Q(-x,a) be two points on the upper half of the ellipse

    [tex] \frac{x^2}{100}+\frac{(y-5)^2}{25}=1 [/tex]

    centered at (0,5). A triangle RST is formed by using the tangent lines to the ellipse at Q and P.

    Show that the area of the triangle is

    [tex]A(x)=-f'(x)[x-\frac{f(x)}{f'(x)}]^2 [/tex]

    where y=f(x) is the function representing the upper half of the ellipse.

    - I know f'(x) is the slope of the tangent line
    - I have the equations for the top half of the parabola as well as the derivation
    - I tried doing slope point form but for some reason it didn't work as I kept getting height = f(x)
  2. jcsd
  3. Dec 13, 2007 #2


    User Avatar
    Homework Helper

    The ellipse appears to be completely irrelevant to the proposition, except for the purpose of imposing a symmetry about the y-axis. I am going to make one small change and call the points on the ellipse (x_0, a) and (-x_0, a).

    I am assuming that the base of the triangle RST is the x-axis, since nothing is given about that in the problem. The tangent lines to the point on the ellipse meet on the y-axis at a point (0, b). Since the slopes on the tangent lines will be +/- f' , the x-intercepts of those tangent lines will be (-b/f' , 0) and (b/f' , 0). Thus the area of the triangle is

    A = (1/2) · (2b/f') · (b) = (b^2)/f' .

    Now the slope of the tangent line from (0, b) to (x_0, a) is

    f' = (a - b)/(x_0) ,

    which is negative, which is also evident from the geometry. We can solve to find

    b = a - (x_0)·f' = f - (x_0)·f' ,

    since a = f(x_0) and I am transferring to the notation of the problem.

    Substituting this into our area formula gives

    A = { [ f - (x_0)·f' ]^2 }/f'

    = { (f'^2)·[ (f/f') - (x_0) ]^2 }/f'

    = f' · [ (f/f') - (x_0) ]^2 .

    Since I have used the half of the triangle in the positive-x half-plane, f' < 0 , so taking the absolute value of this expression to obtain a positive area will introduce a minus sign.

    Nowhere in this does it seem necessary to use any property of the ellipse other than its symmetry about the y-axis.
    Last edited: Dec 13, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook