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From: aegis on 3 Jan 2010 09:31
I have the following graph:
http://i45.tinypic.com/14c5o2u.png

and I am to find paramettric equations for
the curve that consists of all possible positions
of the point P in the image above, using the
angle theta as the parameter. Also note,
the line segment AB is tangent to the larger circle.
(The inner circle is supposed to be that, a circle --
apologies for making it look like an ellipse).

point A can be defined by the following ordered
pair: (a cos t, a sin t) where t stands for theta.

My book says point B is the ordered pair (a sec t, 0)
but I do not see how. The slope of the line segment
OB(where O represents origin) is: (sin t)/(cos t)
or (tan t) and the derivative of (tan t) is (sec^2 t)
and the slope of the line segment AB would then
have slope (sec^2 t) for some t, but this does not
lead me to point B being defined by the ordered pair
(a sec t, 0)

Where am I screwing this up?
From: Lynne Vickson on 3 Jan 2010 13:01
On Jan 3, 6:31 am, aegis <ae...(a)mad.scientist.com> wrote:
> I have the following graph:http://i45.tinypic.com/14c5o2u.png
>
> and I am to find paramettric equations for
> the curve that consists of all possible positions
> of the point P in the image above, using the
> angle theta as the parameter.  Also note,
> the line segment AB is tangent to the larger circle.
> (The inner circle is supposed to be that, a circle --
> apologies for making it look like an ellipse).
>
> point A can be defined by the following ordered
> pair: (a cos t, a sin t) where t stands for theta.
>
> My book says point B is  the ordered pair (a sec t, 0)
> but I do not see how.  The slope of the line segment
> OB(where O represents origin) is: (sin t)/(cos t)
> or (tan t) and the derivative of (tan t) is (sec^2 t)
> and the slope of the line segment AB would then
> have slope (sec^2 t) for some t, but this does not
> lead me to point B being defined by the ordered pair
> (a sec t, 0)
>
> Where am I screwing this up?

Look at the triangle OAB with right angle at point A. If x = |OB| we
have x*cos(t) = a, so x = a/cos(t).

R.G. Vickson
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