Re: Geometry with parabola...

On Apr 30, 5:03 am, "Philippe 92" <nos...@xxxxxxxxxxxx> wrote:
bill a écrit :



On Apr 29, 6:15 am, "Philippe 92" <nos...@xxxxxxxxxxxx> wrote:
bill a écrit :
On Apr 18, 1:49 am, "mina_world" <mina_wo...@xxxxxxxxxxx> wrote:
Hello teacher~

http://board-2.blueweb.co.kr/user/math565/data/math/paramid.jpg
How do you show it ?
Do you use complex algebra to show it ?
or pure geometry ?

Here's a solution using analytical geometry:
Given x = k*y^2. Then,
y = Sqrt(x/k);
sqrt(x) = sqrt(k) * y

The slope of the curve; ie C_m is
C_m = dy/dx= ...

At x=0, dy/dx= infinity...

C_m is also infinity.

Better, never divide dy by dx

How else do you get the slope?

You don't really need the *slope* (defined as dy/dx)
but merely two vector components of the directions, that is
vector (dx, dy), and vector (x2-x1, y2-y1) are parallel (colinear)
if and only if :
dx*(y2-y1) = dy*(x2-x1) whatever may be the signs of the components,
and even when one component (example dx) is 0.

and never extract sqrt when unnecessary.

It was convenient.

See below. It was not so much, because you need to consider separate
cases for the signs (or to use ellipsis : "we could proof in the same
way for other cases")

As I said in a previous post
x = k*y^2 <=/=> sqrt(x) = sqrt(k) * y

If I consider that parabola is the union
of two curves; ie. y = Sqrt(x/k) and
y = -Sqrt(x/k), then sqrt(x) = sqrt(k) * y
is correct.

No, because this implies y >= 0
Definition of sqrt(t) :
the *positive* real number whose square is t

Full parabola is x = k*y^2
You are not allowed to extract square roots here.
You should consider that parabola is the union of the *TWO* curves
y = Sqrt(x/k) *and* y = -Sqrt(x/k)

Shouldn't it be -y = Sqrt(x/k)?

u = -v and -u = v are equivallent, aren't they ?

I did. Otherwise, I could not have gotten a negative y-coordinate for
B.

here is your "mistake" : y_B needs not be negative !

and consider 4 separate cases for A, B on these *TWO* curves.

I can think of only 2 cases; ie, when the
slope of AB is positive and when it is negative.

The slope's sign doesn't really matter, but you need to consider the
cases :
- when A and B are on the same branch (that is y1 and y2 same sign),
- when they are on different branches (y1 and y2 opposite signs)

Oh, I see why you think slope's sign is important : thats another side
effect of your +/- sqrt... : the branch on which lies P !

It is not because on mina's drawing A and B have opposite signs, and
slope is positive, that the property is not valid for *any* A and B !
Your proof is then only for the case when A and B are as in the
drawing, not for *any* A and B, anywhere on the parabola.

Yes, in this specific case you could effectively write
y_A = + sqrt(...) and y_B = - sqrt(...)
but what for other locations of A and B ?





So I just keep the parabola as : x = k*y^2.

Let A : y1 => x1 = k*y1^2
B : y2 => x2 = k*y1^2
with y1 != y2 (otherwise A = B)

Tangent slope at (x,y) defined by :
dx = 2*k*y*dy

Tangent parallel to AB is then defined by :
dx*(y2-y1) = dy*(x2-x1)
that is :
2*k*y*(y2-y1) = k*(y2^2 - y1^2) = k*(y2 + y1)*(y2 - y1)

hence y = (y1+y2)/2.
QED.

I did not seem to me that the question as
posed needed a rigorous formal proof.

There is allways a benefit to search for rigourous proofs.
I agree, this "little exercise" doesn't require so much noise !

Was it wrong for me to post a solution
that was not technically perfect?

Of course not, And I didn't say your proof is not good.
I just say that with your technique (sqrt and slope) you need to
consider separately several cases (or use ellipsis for other cases
you don't effectively write in your proof) :
- the case when x1=x2 (vertical line)
- the case when A and B are on same positive branch
- the case when A and B are on same negative branch
- the case when A and B are on different branches
(and because of the slope's sign, even more cases)

Just because your equations depend on the sign for the sqrt, and
because you consider the slope itself as a real number, and infinity
is not a real number.

My method (with vectors) is valid for all cases at once.
Thats the only reason for my post : to give an other method which
avoids to consider separate cases.

Regards.

--
Philippe C., mail : chephip+n...@xxxxxxx
site :http://chephip.free.fr/ (recreational mathematics)

Thanks for your help and advice. Now if I could only convince myself
that
y = sqrt(x) denies the negative root of x. But that's a different
story.

Bill
.

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