1. ## equidistant problem

Question: A point P(a,b) is equidistant from the y-axis and from the point (4,0). find a relationship between a and b

My attempt
using the formula to find the coordinates on y-axis
$m=[(\frac{x_{1}+x_{2}}{2}),(\frac{y_{1}+y_{2}}{2})]$

find the a
since the coordinate is on the y-axis therefore x= 0
hence
$\frac{0+4}{2} =2 = a$ therefore a = 2

since (4,0) P(a,b)
$\frac{0 + y_{2}}{2} =b$

therefore y=2b
the coordinates on the y-axis is (0,2b) let 's call it b and a= (4,0)

find the distance of BP and AP
$BP = \sqrt{(2b-b)^2 +(0-a)^2}$
$BP = \sqrt{b^2+a^2}$

$AP = \sqrt{(4-a)^2 +(0-b)^2}$
$AP = \sqrt{16-8a+a^2 +b^2}$

find AB
$\sqrt{(4-0)^2+(0-2b)^2}$ $\sqrt{16+4b^2}$

AB = AP + BP
$16+4b^2 = 16-8a+a^2 +b^2 +b^2+a^2$
$b^2 = a^2-4a$

the answer in the book is
$b^2 = 8a -16$
where did i go wrong?

2. ## Re: equidistant problem

let's used squared distance instead of distance for convenience

the squared distance of $(a,b)$ to the y-axis is $a^2$

the squared distance of $(a,b)$ to the point $(4,0)$ is $(a-4)^2 + b^2$

set these two equal

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

$a^2 -8a + 16 + b^2 = a^2$

$b^2 = 8a - 16$

3. ## Re: equidistant problem

Originally Posted by romsek
let's used squared distance instead of distance for convenience

the squared distance of $(a,b)$ to the y-axis is $a^2$

the squared distance of $(a,b)$ to the point $(4,0)$ is $(a-4)^2 + b^2$

set these two equal

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

$a^2 -8a + 16 + b^2 = a^2$

$b^2 = 8a - 16$
I thought this question as this but after your correction for I noticed that the question had this sketch in mind

4. ## Re: equidistant problem

Originally Posted by romsek
let's used squared distance instead of distance for convenience

the squared distance of $(a,b)$ to the y-axis is $a^2$

the squared distance of $(a,b)$ to the point $(4,0)$ is $(a-4)^2 + b^2$

set these two equal

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

$a^2 -8a + 16 + b^2 = a^2$

$b^2 = 8a - 16$
thnks