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wow, im a known associative with a moderator now, score!!!

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(a) is more corrct though from my diagram.

here's how i draw mine carrot when i take my time :P

the locus is \frac{(x-a)^{2}}{b^{2}} + \frac{y^{2}}{b^{2}} = 4 To deduce this we let z=x_{1} + iy_{1} subject to x_{1}^{2} + y_{1}^{2} = 4 for our question a + bz we want to find the locus of this point so we let x= Re(a+bz)=a+bx_1 and we let y=Im(a+bz)=by_{1} then you just make our...

well if they were planning to put it in question 6 as of this year then you just ruined it for them by exposing it :(

i did that originally, but then the question is too trivial, its a much funner exercise than a couple of lines doing it this way

yeh this is pretty decent, cambridge has a funner one of these type of cancellation reduction formula ones, a lot more challenging

better train hard then, this test will not be easy at all

complex numbers curve sketching polynomials conics harder 3u (induction and circle geo) 20 marks each topic, total 100 marks, 12-3pm old building friday 13th goodluck, i bet on you to come around 1st/2nd, dont disappoint, you have a reputation to uphold :P

spiral, holiday test next friday is now 3 hours, 12-3

Yes you are right, I proved it but it was a long process in doing so. Basically I used fact that the three numbers there were real to deduce a relation in terms of theta alpha and beta and then subbed it into abc. After a lot of trig manipulation I showed That the a cancels out I the abc and...

we are given: tanx=ktany the answer has sin only so lets change evrything in terms of sin and cos \frac{sinx}{cosx}=k\frac{siny}{cosy} \\ sinxcosy=ksinycosx \\ \frac{1}{2}(sin(x+y)+sin(x-y)=\frac{k}{2}(-sin(x-y)+sin(x+y)) \\ sin(x+y) + sin(x-y) = -ksin(x-y) + ksin(x+y) \\...

just a side note to the above, for these types of questions it beneficial to work with mod arg form as it has a lot nicer properties than cartesian form, such as just adding args when multiplying. The algebra involved in say letting a = x+iy would be very tedious. So for future questions you...

yes the answer is r^{3} the easiest way to do it is as follows, let a=rcis\theta, \; b=rcis\alpha, \; c=rcis\beta since mods are all equal. Now a+bc = r(cis\theta + cis(\alpha + \beta)) since a + bc is real, Im(a +bc)=0, therefore: Im(a+bc)=r(sin\theta + sin(\alpha+\beta))=0, \\...

Re: 2012 HSC MX2 Marathon damn you carrot

Re: 2012 HSC MX2 Marathon looking at the solution one of my students put forth, not too long

Re: 2012 HSC MX2 Marathon this the tech test?

Re: 2012 HSC MX2 Marathon i really like (b) a lot, ill let the 4u kids have a go first, but if no one posts a solution ill post one tonight