# old ruse question

• 6 Feb 2018, 5:28 PM
altSwift
old ruse question
Attachment 34446

Did they make a mistake on line two, with x(x-a)?? [When I multiplied out, i got x(x+a), ending up with the locus as a circle with centre (-a/2, b/2) instead of (a/2, b/2)]

also, is there supposed to be a restriction on the locus? like excluding (a,0) (because z =/ a on the denominator)

Thanks!
• 6 Feb 2018, 5:47 PM
fan96
Re: old ruse question
Note the denominator in the first line

$\frac{z - ib}{z - a} = \frac{x + iy - ib}{x + iy + a}$

That looks wrong. However they later multiplied by $(x-a)$ instead, so it seems two mistakes have cancelled out?

I don't think there are any further restrictions.

The only conditions given to you are

$\text{Re}\left(\frac{z - ib}{z - a}\right) = 0$

and

$a,b\in\mathbb{R}$
• 6 Feb 2018, 6:05 PM
darkk_blu
Re: old ruse question
$\frac{z - ib}{z - a} = \frac{x + iy - ib}{x + iy + a}$
$Should Have been\frac{z - ib}{z - a} = \frac{x + iy - ib}{x + iy - a}$
in the very first line, its works out as $(\frac{a}{2},\frac{b}{2})$
• 6 Feb 2018, 6:07 PM
altSwift
Re: old ruse question
Quote:

Originally Posted by fan96
Note the denominator in the first line

$\frac{z - ib}{z - a} = \frac{x + iy - ib}{x + iy + a}$

That looks wrong. However they later multiplied by $(x-a)$ instead, so it seems two mistakes have cancelled out?

I don't think there are any further restrictions.

The only conditions given to you are

$\text{Re}\left(\frac{z - ib}{z - a}\right) = 0$

and

$a,b\in\mathbb{R}$

ahh I see it now haha, yeah that confused me for a while.

so would they give a specific condition for z =/ a for some locus question like

$w = \frac{z - ib}{z - a}$
• 6 Feb 2018, 6:11 PM
darkk_blu
Re: old ruse question
I can't believe that question was also in my 4U test today, but I really shouldn't have been. I stuffed up and put $(a,b)$ instead, lol, lost a mark.
• 6 Feb 2018, 6:14 PM
altSwift
Re: old ruse question
Quote:

Originally Posted by darkk_blu
I can't believe that question was also in my 4U test today, but I really shouldn't have been. I stuffed up and put $(a,b)$ instead, lol, lost a mark.

rip, are you from ruse as well?
• 6 Feb 2018, 6:27 PM
fan96
Re: old ruse question
Quote:

Originally Posted by altSwift
ahh I see it now haha, yeah that confused me for a while.

so would they give a specific condition for z =/ a for some locus question like

$w = \frac{z - ib}{z - a}$

I'm not sure what you're asking, sorry.
• 6 Feb 2018, 6:45 PM
darkk_blu
Re: old ruse question
Quote:

Originally Posted by altSwift
rip, are you from ruse as well?

Nah, I'm not that smart, btw do you know if people actually get 100 in ext 2 or is that just scaling?
• 6 Feb 2018, 6:47 PM
camelrider
Re: old ruse question
The graph of a locus represents all the possible points that satisfies a given equation.

You have to exclude (a,0), as it lies on the circle.

Otherwise, you are saying that z=a satisfies Re(z-ib/z-a)=0, which is false.
• 6 Feb 2018, 7:20 PM
altSwift
Re: old ruse question
Quote:

Originally Posted by darkk_blu
Nah, I'm not that smart, btw do you know if people actually get 100 in ext 2 or is that just scaling?

100 as in raw mark? Apparently Anthony Henderson from USYD got 100% in 93'
• 6 Feb 2018, 7:22 PM
altSwift
Re: old ruse question
Quote:

Originally Posted by camelrider
The graph of a locus represents all the possible points that satisfies a given equation.

You have to exclude (a,0), as it lies on the circle.

Otherwise, you are saying that z=a satisfies Re(z-ib/z-a)=0, which is false.

I was thinking this... I wonder why they didn't include it in the answers, thanks!

Quote:

Originally Posted by fan96
I'm not sure what you're asking, sorry.

All good, camel just explained it