Precalculus Problem (Ellipse) Revisited

[mmm4444bot]

Hello everyone:

Now that I've got the diagram to appear, I'm reposting my earlier question.

The problem is to prove (without using calculus) that angles alpha and beta are equal.

My difficulty lies in not knowing how to determine unique angles, since the terminal side of the angle is not a straight ray but the curved line of the ellipse.

~ Mark

 

[tkhunny]

You are going to have to get over the meaningless fact that the angles are drawn to the ellipse and not to the line tangent to the ellipse. The terminal point on the ellipse is just not meaningful, since it is a function of where you draw the little arc-thing. Extend the little arcs to the tangent line and prove that.

 

[mmm4444bot]

Thanks For Your Response

Hello tkhunny:

I appreciate your input. However, I need to say that the angles are defined in the text quite explicitly. Angle beta, for instance, is described as "... the angle formed by [the line segment] PF2 and the ellipse...". (This description matches their diagram.) Nothing is mentioned about the tangent line at P.

Do you suppose that the author made a mistake?

~ Mark

:?

 

[tkhunny]

I'm going to have to argue with the problem statement if it insists that the terminal side is actually ON the ellipse. As stated before, one must specify the distance from the point P in order for that to be meaningful. If the alpha-arc is closer than the beta-arc, as it appears in your drawing, I think the proof is trivial that the two angles are NOT equal.

The crux of the problem is in the definition of what is meant by an angle, using a line segment and an ellipse, with their intersection being the coterminal point. Check your definition of "angle". I hope it is defined as "...two RAYS...". This thoroughly discards the idea that the terminal side can be a finite piece of an ellipse. It must be the angle formed by Ray(PF) and Ray(half the tangent line).

My views. I welcome others'.

 

[pka]

There should be no argument about the terminology.
This is a well-known theorem:
The focal radii to any point on the ellipse form congruent angles with the tangent at the point.
In the given figure PF<SUB>1</SUB> and PF<SUB>2</SUB> are focal radii.
Therefore TKH is correct and your text has played loose with terminology.
This theorem is an algebraic nightmare to prove without calculus
However here are some facts that may help you.

Let \(\displaystyle c = \sqrt {a^2 - b^2 }\); then \(\displaystyle F_1 - c,0){\rm{ and }}F_2 c,0)\).

The slope of \(\displaystyle F_1 P{\rm{ is }}\frac{{{\rm{y}}_{\rm{1}} }}{{x_1 + c}}\) and slope of \(\displaystyle F_2 P{\rm{ is }}\frac{{{\rm{y}}_{\rm{1}} }}{{x_1 - c}}\).

Now the equation of the tangent is \(\displaystyle y = mx + \sqrt {a^2 m^2 + b^2 }\) where m is its slope.

Recall that the angle between two lines with slopes m<SUB>1</SUB> and m<SUB>2</SUB> is \(\displaystyle \theta = \arctan \left( {\frac{{m_1 - m_2 }}{{1 + m_1 m_2 }}} \right)\).

Now the problem is to show that those two angles are equivalent.
I certainly would NOT even try the algebra.

 

[mmm4444bot]

Thanks Again.

tkhunny wrote:

> Check your definition of "angle".
> I hope it is defined as "...two RAYS..."

Exactly. Thus my posted perplexity.

Using the text's description (and confirming diagram), there is no way to determine angles alpha and beta.

I don't have time to check pka's response right now, but I will do so later. The vast majority of the exercises in this text are fairly basic, so I'm not sure whether I need to experience a "nightmare" of algebra.

I thank you both for your responses.

~ Mark

 

[mmm4444bot]

I'm getting there ...

pka wrote:

"... your text has played loose with terminology."

I feel that pka is being too kind to the authors. There is no gray area in what is written; it is just plain wrong.

Anyway, back to the issue at hand, I am getting closer to proving that the corrected angles alpha and beta are equal.

Awhile back, I was using a paper circle exercise which involved marking a point on a circle (other than at the center), and then choosing multiple points on the circumference which were then brought to touch the point on the circle by folding. After several folds, the creases intersected to form the shape of an ellipse.

Yesterday, I was working on a problem that involved a smaller circle rolling around inside of a larger circle. Then it occurred to me that I could position an ellipse inside the larger circle such that a fold could be made where the crease would be tangent, and the point of tangency could be the center of a smaller circle tangent to the larger circle.

I drew several ellipses and circles last night, and in each case I was able to come within one step of proving that angles alpha and beta are equal. Now, I just need help with the last step.

Here is my diagram:

Points F1 and F2 are the foci.

Point T is the intersection of tangent line L and the segment QF2.

Line L is the crease made by folding the large circle to bring point Q onto point F2.

Since L is also the diameter of circle C', the distances QT and TF2 must be equal.

So, we have triangles TPF2 and TPQ whose corresponding sides are all equal. This means that angles TPF2 and TPQ are equal.

Line L intersects segment QF1 at point P. Hence, opposite angles SPF1 and TPQ are equal.

This shows that angles TPF2 (alpha) and SPF1 (beta) must also be equal.

The one step that I have not yet worked out is showing that points Q, P, and F1 are always colinear. I'm fairly sure that they are because in each of my careful diagrams I extended segment QP and each time it came very close to F1. (Close enough to lead me to believe that QF1 is the radius of C.)

Any thoughts about this last step anyone?

~ Mark