Given that P is the vertex of a right angled triangle with hypotenuse
AB and catheti parallel to the axes. Write the equation for the
location of the points P for the half-ray r originating at O and
intersecting A and B on the circumferences x^2+y^2=16 and x^2+y^2=25.

I'm an italian high school student having difficulty with my homework
assignment.Please can you explain all steps clearly. I know the answer
is an eclipse but I'm unable to show how !
Thanks Alex

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Request for Question Clarification by richard-ga on 04 Feb 2006 10:03 PST

Catheti (I had to look it up!) are the legs of a right triangle, and
the question is just telling us the triangle is standing up in the
usual way - - its bottom leg is parallel to the x-axis [actually I
think it is sitting on the x-axis, because of the question's reference
to 'O' the origin), and the side leg is parallel to the y-axis (again
I think it is sitting on the y-axis).

As I'm sure you know, x^2+y^2=16 and x^2+y^2=25 are two circles
centered on the origin.  I think that business about a half-ray means
that one triangle leg is 4 units long and the other is 5 units long,
so it's a 3-4-5 right triangle.

So I think they're asking you to fit a continuous curve (yes, it would
be an ellipse) that runs from one circle's circumference (starting at
point (0, 5) if the longer leg is pointing up)and running into the
other circle's circumference at point (4,0), then on to (0, -5) and
(-4, 0)

Shame on them for making it as much an English test as a Math test!

If this is a satisfactory answer for your needs, let me know if I can
post it as an Answer.

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Clarification of Question by jamano-ga on 05 Feb 2006 04:16 PST

So what's the equation for the curve that point P (the vertex of the
triangle), will draw as the line segment rotates along the
circumferences?

NB I had only guessed it was going to be an eclipse but it wasn't
actually given data in the question. I need to show all steps to get
the equation.

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Request for Question Clarification by richard-ga on 05 Feb 2006 07:48 PST

The Google Answers policy about homework questions, as I understand
it, is that we can help or teach, but not do people's work for them.

My interpretation of your geometry problem is that you are being asked
to visualize the two concentric circles (radius 4 and radius 3), and
the two possible 3-4-5 right triangles that can sit inside those
circles (one triangle is base 4 and height 3, the other is base 3 and
height 4).

Pick one of those triangles, say base 3 height 4, and you'll see its
hypotenuse meets the big circle at point (0,4) and the smaller circle
at point (3,0).

I read the question as asking you to come up with the formula, thank
you Researcher Mathtalk-ga for these words, for the locus of points P
that are the vertices of a right triangle opposite hypotenuse AB with
"legs" (catheti) parallel to the coordinate axes.

Why that locus describes an ellipse, and what the algebraic formula is
for that ellipse, I'll leave for you to figure out.

Since I'm not completely answering your question, I'll leave this
here, for free, rather than post it as an answer and be paid.

-R

I noticed that you have not received an answer for your homework question.
A homework help site such as www.eacademichelp.com specializes in
homework assistance. Hope this helps.

I think what you are talking about is best understood by looking at
the parametric equations of an ellipse.

x = a cos(t)
y = b sin(t)

a and b are the major and minor axes of the ellipse.
a = 5 and b = 4

Note that (x/a)^2 + (y/b)^2 = cos^2(t) + sin^2(t) = 1
You should recognize this as the equation of an ellipse.

Put an arbitrary point P on the outer circle.  Draw the line segment
OP.  Note where it intersects the inner circle.  Construct a vertical
line thru point P and a horizontal line thru where the OP intersects
the inner circle.  The intersection of the horizontal and vertical
lines (let's call that point Q) so constructed will be a point on the
ellipse.  As you drag point P around the outer circle, Q will follow
the path of an ellipse.

If you have access to Geometer's Sketchpad, that would be a handy tool
to create this construction dynamically.

I should have added, as a final bit of clarity, that the point Q can
be easily seen to be described by the parametric equations given in my
previous posting.

Jamano and Richard,

First of all, please note that the two sides of the triangle you have
determined, namely 4 and 5, are the catheti.  This missing side is the
hypotenuse.  So its length is not 3 but rather sqrt(4^2+5^2) or
sqrt(41).  However, the length of the hypotenuse is not needed to
solve the problem.

The question is not precisely worded.  However, it seem that what you
are looking for is the locus of points that intersect the two circles
on the two axes.  The triangles direct you to 4 of the points.  There
are two solutions.  I directed you to how to find one of the solutions
in my first post.

The points of intersection are either:

(5,0) (0,4) (-5,0) (0,-4) :this was the first post
or
(4,0) (0,5) (-4,0) (0,-5)

In the first case the major axis is along the x axis, and in the
second case it is along the y axis.  My first post directed you to the
first solution.  If you switch the values of a and b you will get the
second solution.

In my first post, the point Q traces out the path of the ellipse. 
Please note also, that I gave you the generic formula for the equation
of an ellipse centered on the origin, which is what you want here. 
With the appropriate switching of construction for the horizontal and
vertical lines, you can construct the second point Q that will trace
the path of an ellipse with the major axis being along the y axis.

Please let me know if this helps or you still have questions.

The above comment rightly points out that the legs of the triangle are
4 and 5.  Funny how I fell into picturing a 3-4-5 triangle when a
4-5-sqrt(41) triangle was wanted.
-R

To echo what ansel001-ga has said, there are two ellipses.  Given the
ray from the origin O intercepting the inner circle at A and the outer
circle at B (the wording of the original doesn't really specify which
is which), there are two points P which form a right triangle with
hypotenuse AB and "legs" parallel to the coordinate axes:

 - Use the x coordinate of A and the y coordinate of B

 - Use the x coordinate of B and the y coordinate of A

If the two cases for P are treated separately, each gives rise to an ellipse.

The parametric approach given by ansel001-ga makes short work of
finding the equation for either of the ellipses.  However it might be
beyond what methods are intended for the high-school assignment.  I'd
consult with the instructor to see if it is allowed, or if the
assignment requires a derivation using only Cartisian coordinates.  If
it is the latter, the difficulty becomes a matter of developing an
equation that relates the the x coordinate of A with the y coordinate
of B (and vice versa for the other ellipse) using only the equations
of the circles and the fact that A,B lie in a line with the origin.


regards, mathtalk-ga

Mathtalk,

You may have a point about the parametric equations.  I don't remember
if I learned about them in high school or college.  When are they
usually introduced?

Hi, ansel001-ga:

Your Comments on this thread have been thoughtful and well written.

I think parametric equations are covered in high school, e.g. the
parametric equation of a line or a circle.  I'm just suggesting a
classroom exercise might have been intended to emphasize either
geometric or polynomial tools.  It's nice to be able to see a
calculation worked out with more than one perspective!

regards, mt

Thanks Mathtalk.