- #1

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Any hints on how to go about this appreciated.

Thanks.

- Thread starter MASH4077
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- #1

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Any hints on how to go about this appreciated.

Thanks.

- #2

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Write down the appropriate equation. Expand, simplify.

- #3

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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

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- #5

HallsofIvy

Science Advisor

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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:

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.

[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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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

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Isn't this a simple parabola with the focus at (4,0) and the directrix the Y-axis?

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