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
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!
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
Last edited by fan96; 6 Feb 2018 at 5:53 PM.
in the very first line, its works out as
Last edited by darkk_blu; 6 Feb 2018 at 6:08 PM.
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.
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.
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