Geometry Question

  • Thread starter MASH4077
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  • #1
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A point P(a, b) is equidistant from the y-axis and from the point (4, 0). Find a relationship between a and b.

Any hints on how to go about this appreciated.

Thanks.
 

Answers and Replies

  • #2
1,444
4
Write down the appropriate equation. Expand, simplify.
 
  • #3
12
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Hi,

I've tried numerous ways of tackling this but I can't seem to get the answer that I have in my solutions booklet. Looking to see if anyone can give an alternative starting. Anyway here's one approach I used...

Let Q be the point the line A(4, 0) -> P(a, b) cuts the y-axis.

[tex]AP^2 = (4 - a)^2 + b^2[/tex]

[tex]PQ^2 = a^2 + (b - (4b/(4-a)))^2[/tex]

but [tex]AP^2=PQ^2[/tex]. In trying to simplify that I get something that isn't even close... which is:

[tex](b^2(6-a))/((4-a)^2) = 2 - a[/tex]

Thanks.
 
  • #4
1,444
4
Take several points P(a,b), draw them on a piece of paper, and for each one think what is its "distance from the y-axis". Draw this distance. Perhaps you will notice something?
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
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Hi,

I've tried numerous ways of tackling this but I can't seem to get the answer that I have in my solutions booklet. Looking to see if anyone can give an alternative starting. Anyway here's one approach I used...

Let Q be the point the line A(4, 0) -> P(a, b) cuts the y-axis.

[tex]AP^2 = (4 - a)^2 + b^2[/tex]

[tex]PQ^2 = a^2 + (b - (4b/(4-a)))^2[/tex]

but [tex]AP^2=PQ^2[/tex]. In trying to simplify that I get something that isn't even close... which is:

[tex](b^2(6-a))/((4-a)^2) = 2 - a[/tex]

Thanks.
That's the basic idea but I see no reason to introduce "Q". The distance from (a, b) to the y-axis is the distance from (a, b) to (0, b) and is equal to |a|. The distance from (a, b) to (4, 0) is [itex]\sqrt{(a- 4)^2+ b^2}[/itex]. (a, b) is "equidistant from the y-axis and (4, 0)" if and only if those are equal:
[tex]|a|= \sqrt{(a- 4)^2+ b^2}[/itex]
to get rid of the square root on the right, square both sides. Fortunately [itex]|a|^2= a^2[/itex] so that also gets rid of the absolute value:
[tex]a^2= (a- 4)^2+ b^2[/tex]
Multiplying out [itex](a- 4)^2[/itex] will also give an [itex]a^2[/itex] on the right which will cancel the one on the left. You will have an equation with a and [itex]b^2[/itex] but no [itex]a^2[/itex]- a parabola.

Personally, I would have used (x, y) rather than (a, b) but its the same thing.
 
  • #6
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Ah, I see...

That makes perfect sense. The problem I had was always ending up with an a^2 after doing all the algebraic manipulation.

Arkajad and HallsofIvy, many thanks. :)
 
  • #7
coolul007
Gold Member
265
7
Isn't this a simple parabola with the focus at (4,0) and the directrix the Y-axis?
 

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