hi, all,

im a 6th form further maths a-level student, and we've done loads about reflecting a line, (point, shape etc.) in a straight line, (eg. y=x), but im intrigued about how to reflect a line in a curve, (eg. y=x^2). i have a theory that you just draw a normal to the curve joining a series of points on the line being reflected, going equidistant to the other side of the curve.

(i hope this makes sense, cos i'm predominantly a mathematician, and can't speak or write properly!)

all the best ian

Right, this is what you do. From every point of your shape draw a perpendicular to the curve and extend it to twice its length.

One question I have is the reason you would like to do that. If, for example, you wish to figure out your appearance in a crooked mirror, the construction will be different.

All the best,
Alexander Bogomolny

LAST EDITED ON Feb-02-01 AT 00:18 AM (EST)

If I may modify my previous post, the situation is probably a little more complex than that.

Instead of saying "Right, this is what you do," I should have said "You are within your rights to do that." Yours may not be the only possible definition of the reflection in a curve. The well known "reflection" in a circle is inversion.

Two points P and Q are inversions of each other in a circle with center O and radius r iff the three points O, P, Q are colinear and OP·OQ = r2.

Thanks for your help. i can't really answer your question, why i'd want to, i guess it's just for completeness. it annoys me when i don't know how to do something that interests me.

my problem now is how tofind the equation of the new line. i realise this is probably a bit too much to go into easily, but do you know where i can find it out?

thanks again
ian

>Thanks for your help. i
>can't really answer your question,
>why i'd want to, i
>guess it's just for completeness.
> it annoys me when
>i don't know how to
>do something that interests me.

What I asked was, "What is it you see in your mind's eye when you talk about those reflections?" The reason I asked was because there may be several kinds of reflections - that's all.

>my problem now is how to find
>the equation of the new
>line. i realise this
>is probably a bit too
>much to go into easily,
>but do you know where
>i can find it out?
>

It's pure analytic geometry and/or some Calculus. The most essential part of you query is being able to find the point on a given curve nearest to a given point.

Say, you are given a point (x0, y0) whose reflection in y = x2 you intend to find. You'll have to minimize the sqaure of the distance:

d(x) = (x - x0)2 + (x2 - y0)2

Equate the derivative to 0 and see if you can solve the equation. If you can, assume x1 is a solution. Then the reflection point is given by

(x, y) = 2(x1, y1) - (x0, y0)

hi alex,
you're probably bored of me by now, but i've been thinking again.
please could you see if this seems reasonable. i realise it's probably difficult to understand, without the graphs (sorry).

consider the set of axis "y" against "f(x)".

the line "y=f(x)" is a straight line through the origin.

the line "y=x" is the inverse of f(x). i.e. "y=(f^-1)(f(x))".

reflect this in the 'straight line' "y=f(x)" and you get
"y=f(f(x))", i think

this means that y=x reflected in y=f(x) gives y=ff(x)!

i realise that there's a problem with,for example f(x)=x^2 but i know that this comes from the squaring. it can be solved by reflecting y=mod(x)
im trying to do the same for y=g(x) reflected in y=f(x), but im not sure. at the moment i think its y=fg(-1)f(x).

sorry to bother you again, but thanx for your help. none of my tutors bother to help me.

thanx!

>consider the set of axis "y"
>against "f(x)".

The function f(x) must be monotone for that to make sense.
>
>the line "y=f(x)" is a straight
>line through the origin.

Yes. It's the first quadrant's diagonal.
>
>the line "y=x" is the inverse
>of f(x). i.e. "y=(f^-1)(f(x))".

Yes.

>reflect this in the 'straight line'
>"y=f(x)" and you get
>"y=f(f(x))", i think

Yes. That's true.

>this means that y=x reflected in
>y=f(x) gives y=ff(x)!

That's right, but why the exclamation mark?

>i realise that there's a problem
>with,for example f(x)=x^2 but i
>know that this comes from
>the squaring. it can
>be solved by reflecting y=mod(x)

That I do not know.

>sorry to bother you again, but
>thanx for your help.
>none of my tutors bother
>to help me.

So, what are they for?