This will be my first 'original work' post (NOTE: Chances are, this is not undiscovered knowledge. Somebody, somewhere must have already discovered this. However, I am not aware of any work already done on the problem, and all work has been done independently ie. I worked this out, I didn't copy it from somewhere else). Hope you guys like it :)
I was on a plane back from South Africa (which is an AMAZING place to visit if you ever get the chance) and anxiously awaiting GCSE results. To take my mind of the impending judgement which, on the basis of a poorly designed exams, would give me a grade to decide my aptitude for the given subject for the rest of my life... (-_- ugh...GCSEs)
I started thinking about Maths (as you do) and the problem that had been bugging me for a while. If you draw a curve:
y= f(x)
and a point
(x1 , y1)
you should be able to reflect the point in the curve such that the line connecting the point and the reflected point is a normal (perpendicular) to the curve (what I like to call a 'proper' reflection) like so:
Well, this is the equation I came up with:
f(x) + [(x1 + x)/f '(x)] – y1 =
0
The solutions (values of x which satisfy the above) to the first order differential equation above can be input into the following formula to give the set of 'proper' reflection points:
(2x – x1 , 2 f(x) – y1)
So, there we are. Sorry if this was a bit confusing, blogger is a bit rubbish for equations and diagrams. Please comment with any difficulties in understanding or general questions and I'll do my utmost to answer. If you have some free time, why not try to see how I proved this solution. The proof of this is unfortunately far too large to contain in this margin (hehehe, sorry, couldn't resist ;) ). Not difficult really, but interesting and with enough pitfalls to dodge to keep you on your toes.
If you're wondering whether to comment or not... please comment :p
Thanks for reading :)
Anand
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