Hi i am french and i want an help on my math exercise (math are the same thing in every language), thank you for your good will.

below my exercise and my beginig of reponse

[tex]5.On considers the set arc[tex] \ gamma [/ tex] defined by: [tex] \ vec (OM (t) = R) (\ frac (1-t ^ 2) (1 + t ^ 2) \ vec (i) + \ frac (2t) (1 + t ^ 2) \ vec (j) [/ tex] with t [tex] \ in [/ tex] R.

It should be noted M (l) the point of setting the parameters of the arc.

6.Soit t u 2 real. Check that: [tex] (1-p) x + sy-R (1 + p) = 0 [/ tex] is an equation of the line (M (L) M (u)), if we asked s = l + u and p = l * u.

If t = u, the notation (M (t) M (u)) will designate this time the tangent in M (t) [tex] \ gamma [/ tex].

We admit, without the check, the equation is still found in this case. Give the equation.

7.a) Let M point [tex] \ gamma [/ tex] parameter l. Show that, except in one case to clarify its symmetrical orthogonal M 'compared to (O, [tex] \ vec (j) [/ tex]) is a point of [tex] \ gamma [/ tex]. Check that the parameter M 'is [tex] l' = \ frac (1) (t) [/ tex].

If [text] a_0 [/ tex] denotes the geographical coordinates (R, 2R), show that when the [tex] neq [/ tex] 1, the right ([text] a_0 [/ tex] M) cuts [tex] \ gamma [/ tex] to the point of setting [tex] \ frac (1) (1)-t [/ tex].

(You can use the résutats 6).

My answers (for now):

2. I get the equation: (1-lu) x + (P + u) yR (1 + lu) as equation 1 is indeed the right equation (M (L) M (u))

-The equation of the line (M (t) M (u)) is the same: (1-lt) x (t) + (P + t) y (t)-R (1 + lt) (1) .

3.a) The symmetric M \ gamma compared to (o, j) is such that: x (-t) =-x (t) and (-t) = y (t).

hence replacing in (1): (lt-1) x (t) + (P + t) y (t)-R (1 + lt) (2): equation of the tangent (M '(t) M (u)) tangent M (t): but how this would he say that my point of \ Gamma, the more I can not find the particular case for which M '\ notin $\displaystyle \ gamma$ , thank you help me please.I use for math latex.