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easy complex numbers question (1 Viewer)

conics2008

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Its simple but cant seem to get my head around it.

Also Conics, havn't done them for so long.
 
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undalay

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For complex

A->B = O->B - O->A
and A->C = A->Bcis45 x 2
O->C = O->A + A->C
 

vds700

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conics2008 said:
Its simple but cant seem to get my head around it.

Also Conics, havn't done them for so long.
For conics:

(i) i assume you're right with this part.

(ii) use perpendicular distance of a point from a line formula, its not that difficult. Let us know if u need the working.

Just a bit of algebra bashing!
 
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tommykins

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vds - I got it down to

a²sin²@ - 2a²bsin@ + ab²
--------------------------------
b²cos²@ + a²sin²@

Wut now?
 

undalay

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erm u did something wrong tommyy ? not sure how u got that
 

tommykins

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Probably, I don't even know if I have the formula correct

(ax + by + c)/sqrt[a²+b²] amirite? (obv aboslute value).

x = +-ae
y = 0
a = bcos@
b = asin@
c = -ab

Is that correct? If so, then probably some algebra error in my working.
 

conics2008

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hey thanks.. just the last step...

I knew we had to use b^2=a^2(1-e^2) and the ellipse..

Thanks Undalay

hey can you explain why AC = 2ABcis45

how can you assume it rotation of 45 ??
 
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duy.le

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oh i dont no if u can do it yet but ill post up my answer cause it did as for a complex number representation. i didnt exactly go through the working out line by line but the general idea is there, please forgive me for any mistakes.

sorry if it doesnt work. im sorta new to the zip thing. (not pro at comps.) :(
 

undalay

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ur rotating A->B

so the tail must be A.
IT must be A->C

it cant be O->C
 

duy.le

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undalay said:
ur rotating A->B

so the tail must be A.
IT must be A->C

it cant be O->C
oh yeah my bad, your right undalay the vector is ac not oc, please note that. :shy: i would of lost one mark or the markers would of overlooked it. hope i dont make that mistake in the hsc.
 

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