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old ruse question (1 Viewer)

altSwift

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ruse-solution.png

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!
 

fan96

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Note the denominator in the first line



That looks wrong. However they later multiplied by 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



and

 
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altSwift

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Note the denominator in the first line



That looks wrong. However they later multiplied by 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



and

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

 

darkk_blu

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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 instead, lol, lost a mark.
 

fan96

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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

I'm not sure what you're asking, sorry.
 

camelrider

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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.
 

altSwift

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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!

I'm not sure what you're asking, sorry.
All good, camel just explained it
 

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