leehuan's All-Levels-Of-Maths SOS thread (2 Viewers)

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InteGrand

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Followup: Obviously arctan(x)+arccot(x) becomes a piecewise function. How do I justify that this isn't the case for arcsin(x)+arccos(x)? Does it basically just have to do with the fact arccot(x) is undefined for x=0 but continuous everywhere else?
 

Paradoxica

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Well, I suppose now is a better time than never to crack down this question of mine...

Take an arbitrary ellipse, centred at the origin and aligned with the cartesian axes for ease of computation.
Construct the director circle of the ellipse.
Take a point on the director circle and construct the obviously perpendicular tangents to the ellipse.
At the two points of osculation, construct normals to the tangents.
What is the algebraic equation that defines the locus of all possible points that are the intersection of these normals?

Repeat the above for the director circle of a hyperbola. Obviously |a|>|b| or else the director circle has imaginary radius, and we are not wishing to deal with 4-dimensional algebraic curves.
 
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seanieg89

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Well, I suppose now is a better time than never to crack down this question of mine...

Take an arbitrary ellipse, centred at the origin and aligned with the cartesian axes for ease of computation.
Construct the director circle of the ellipse.
Take a point on the director circle and construct the obviously perpendicular tangents to the ellipse.
At the two points of osculation, construct normals to the tangents.
What is the algebraic equation that defines the locus of all possible points that are the intersection of these normals?
Do you know for a fact that a nice cartesian form exists? Or will a parametric description suffice?

It is not hard to obtain a parametric formula. (MX2 level conics.)

My method:
1. Show that the director circle has radius sqrt(a^2+b^2) for an ellipse in the standard form.
2. For a point X on the director circle let the two points of contact with the ellipse be A,B and let the normals at A,B meet at N. Then AXBN is a rectangle, and we can obtain N by reflecting X about the midpoint M of AB.
3. We can find the midpoint M of AB in terms of the coordinates x_0,y_0 of X by equating the chord of contact with the cartesian equation of ellipse (ignoring vertical line issues which can be dealt with separately if you like). We do not need to solve the resulting quadratic, we only need to halve the sum of the roots to obtain our midpoint.
4. Our point N then has coordinates given by x_2=x_0+2(x_1-x_0) and similarly for y_2. (Here M=(x_1,y_1).)

My calculation led to



By setting we then obtain a parametric (polar in fact) equation for the locus of N.

With some manipulations, you might be able to find an implicit cartesian equation for N, but I will not spend any time on trying to do this right now.

Out of interest, the parametric plot looks like (for a=sqrt(2),b=1):




If anyone does the same computation, let me know if your parametric plot matches mine, as I was not particularly careful and will not have time to check over this properly till after a meeting tomorrow arvo.
 
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seanieg89

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Equally useful would be a visual confirmation of the above using Geogebra. I will do this tomorrow if no-one else does.
 

leehuan

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MX2 difficulty:

This question got asked whilst I was busy, and was also solved. But I'm asking it here because I'm wondering if anyone knows what the most elegant method of proof is:

 
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