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Finding Dimensions of Elliptical Shadow Cast on Plane

Date: 01/20/2005 at 03:33:38
From: Bertram
Subject: cone - plane intersection resulting in an ellipse

Dear Dr. Math,

A sphere is placed between a X-ray (point) source and an X-ray plate. 
The line from the X-ray source through the center of the sphere is 
perpendicular to the plane of the X-ray plate in this first situation.
The distance between source and plate is known (1200 mm), as well as 
the distance between the source and the center of the sphere (1000 
mm).  The diameter of the sphere is 28 mm.

Now the sphere is shifted 500 mm away from its original position in a 
plane which runs parallel to the plane of the X-ray plate.  So, the 
shortest line between the center of the sphere and the X-ray plate 
still has the same length (200 mm).

Our question is (given the above situation):

How do we calculate the major and minor axes of the resulting 
projected ellipse on the X-ray plate, as well as its surface area?
We are limited in our mathematical knowledge and hope that you can
bring clarity in this matter.

Kind regards!


Date: 01/20/2005 at 19:28:44
From: Doctor Peterson
Subject: Re: cone - plane intersection resulting in an ellipse

Hi, Bertram.

Thanks for writing to Dr. Math.

I have worked out formula for the semiaxes of the ellipse, given the 
following figure, which shows two views of the setup as I understand 
it, with the source at A and the image at PQ:


If we define

  x = horizontal shift DC = 500 mm
  y = vertical distance from source AD = 1000 mm
  h = distance from source to plate AB = 1200 mm
  r = radius of sphere CS = 14 mm

then I find, using similar triangles in each view shown,

  a = PQ/2 = hrsqrt(x^2 + y^2 - r^2)/(y^2 - r^2)

  b = P'Q'/2 = hr/sqrt(y^2 - r^2)

I checked these formulas against the foci found by projecting the top
and bottom of the sphere, and it worked out.  Let me know how this
works for you.  I can show you how I derived these formulas if you are

- Doctor Peterson, The Math Forum 

Date: 01/25/2005 at 11:22:53
From: Bertram
Subject: cone - plane intersection resulting in an ellipse

Dear Dr. Peterson,

Thank you for the prompt response.  This really helps a lot.  Indeed 
it would be even more insightful if you could show me the derivation 
of the formulas.  I have tried it myself, but I don't seem to end up 
with the same results.

Kind regards!


Date: 01/25/2005 at 22:54:32
From: Doctor Peterson
Subject: Re: cone - plane intersection resulting in an ellipse

Hi, Bertram.

Okay, let's go through my derivation, first for the major axis PQ
(using the figure I drew previously):

I didn't label the point of intersection of segments AP and CD; let's
call that E.  Then triangles ADE and CRE are similar, with

  AD = y             CR = r
  DE = u             RE = v

So v/u = r/y, and v = ur/y.  But by Pythagoras,

  r^2 + v^2 = (x-u)^2


  r^2 + (ur/y)^2 = (x-u)^2

Expanding, and multiplying by y^2, we have

  r^2 y^2 + u^2 r^2 = x^2 y^2 - 2uxy^2 + u^2 y^2

Treating this as a quadratic in the unknown u,

  (y^2 - r^2)u^2 - (2x y^2)u + (x^2 - r^2)y^2 = 0

By the quadratic formula,

      2x y^2 +- sqrt[4x^2 y^4 - 4(y^2 - r^2)(x^2 - r^2)y^2]
  u = -----------------------------------------------------
                       2(y^2 - r^2)

      x y^2 +- y sqrt[x^2 y^2 - x^2 y^2 + r^2 y^2 + r^2 x^2 - r^4]
    = ------------------------------------------------------------
                             y^2 - r^2

      x y^2 +- ry sqrt[y^2 + x^2 - r^2]
    = ---------------------------------
                  y^2 - r^2

Now, if you replace E with E', the intersection of AQ and CD 
(extended), you would find the same quadratic equation, so that the
two solutions we have here in fact give the horizontal distances from
D to both E and E', and so the distance from E to E' is the 

           2ry sqrt[y^2 + x^2 - r^2]
   u'- u = -------------------------
                  y^2 - r^2

But the major axis PQ satisfies PQ/EE' = h/y, so the minor semiaxis
is, as I said,

         rh sqrt[x^2 + y^2 - r^2]
  PQ/2 = ------------------------
                y^2 - r^2

For the minor semiaxis b = B'P', consider similar triangles A'R'C and
A'B'P', which give the proportion

  r/b = y/A'P'

and since A'P' = sqrt[b^2 + h^2], we get

  r sqrt[b^2 + h^2] = yb

  r^2 (b^2 + h^2) = y^2 b^2

  r^2 h^2 = (y^2 - r^2)b^2

  b = rh/sqrt[y^2 - r^2]

That completes the derivation.

- Doctor Peterson, The Math Forum 

Date: 02/02/2005 at 11:02:15
From: Bertram
Subject: Thank you (cone - plane intersection resulting in an ellipse)

Dear Dr. Peterson,

Thank you very much for the clear derivation, and for all of your help
with this problem.  Your service is excellent!

Kind regards,

Associated Topics:
College Higher-Dimensional Geometry
High School Higher-Dimensional Geometry

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