Honors Maths Question Help...

Citizen_Erased

Hi all,
Just been getting down to some study and I have come across this question which had me slightly stumped so I went onto the e-xamit site to have a look at their solution.
The Question is LC 2005 , Paper 1 , Question 6, C (iii). You are being asked to prove that the curve is its own image under central symmetry through the point of intersection of the asymptotes .Here is the solution from e-xamit.ie (btw code is cbmoqg if you wanna check it for yourself) and it is fairly comprehensible until the bit I have highlighted in red. Surely I am not mistaking in saying that what it says there is completely incorrect? Any suggestions at all?

Thanks
C_E

no, its correct. they convert the 2 into 2(a-1/a-1). i.e (2a-2)/a-1.

This puts 2a-2-a above the line (a-1 below) which becomes a-2/a-1


make sense?

Citizen_Erased

Yes I have it at a-2/a-1 , what I dont get is how it then jumps it to 2-a/a-1 .

ok

2 - a/(a-1)

becomes

2*(1) - a/(a-1)

becomes 2*(a-1)/(a-1) - a/(a-1)

becomes (2a-2)/(a-1) - a/(a-1)

= (2a-2-a)/(a-1)

= (a-2)/(a-1)

clearer?

oops, yeah that's wrong...

Citizen_Erased

Sorry I am not being very clear here , I have that bit that you have described that is not my problem , hang on a minute I will post image to clarify.
Heres the image ...


{edit} Yep , Emmet theres defo something up with it , sorry I wasn't clearer earlier but it's great to here it from someone else now too. Thanks for the help.

no i see it know, it should have become 2-a/1-a if they were going to multiply it by -1/-1

Citizen_Erased

Actually I think I have it now , it is just a typing error or whatever because later on in the question it is refared to as 2-a/1-a so they did just multiply by -1/-1 but just a typo or whatever in that line and the next is the problem.
So problem solved , thanks for your help Emmet.

C_E