Geometry Question

[MASH4077]

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.

 

[arkajad]

Write down the appropriate equation. Expand, simplify.

 

[MASH4077]

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.

$$AP^2 = (4 - a)^2 + b^2$$

$$PQ^2 = a^2 + (b - (4b/(4-a)))^2$$

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

$$(b^2(6-a))/((4-a)^2) = 2 - a$$

Thanks.

 

[arkajad]

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?

 

[HallsofIvy]

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.

$$AP^2 = (4 - a)^2 + b^2$$

$$PQ^2 = a^2 + (b - (4b/(4-a)))^2$$

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

$$(b^2(6-a))/((4-a)^2) = 2 - a$$

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 $\sqrt{(a- 4)^2+ b^2}$. (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 $|a|^2= a^2$ so that also gets rid of the absolute value:
$$a^2= (a- 4)^2+ b^2$$
Multiplying out $(a- 4)^2$ will also give an $a^2$ on the right which will cancel the one on the left. You will have an equation with a and $b^2$ but no $a^2$- a parabola.

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

 

[MASH4077]

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. :)

 

[coolul007]

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