# Hyperbola and dot product

## Homework Statement

A point P moves so that its distances from A(a, 0), A'(-a, 0), B(b, 0) B'(-b, 0) are related by the equation AP.PA'=BP.PB'. Show that the locus of P is a hyperbola and find the equations of its asymptotes.

## The Attempt at a Solution

AP.PA' = $((a-x)\boldsymbol i + y\boldsymbol j).((-a-x)\boldsymbol i+y\boldsymbol j)$
AP.PA' = $x^2-a^2+y^2$
BP.PB' = $((b-x)\boldsymbol i + y\boldsymbol j).((-b-x)\boldsymbol i+y\boldsymbol j)$
BP.PB'= $x^2-b^2+y^2$

So

$a^2=b^2$

This result sugests that their is no constraint on P. This is not consistent with the question.

Last edited:

## Answers and Replies

andrewkirk
Homework Helper
Gold Member
AP.PA' = $((a-x)\boldsymbol i + y\boldsymbol j).((-a-x)\boldsymbol i+y\boldsymbol j)$
AP.PA' = $x^2-a^2+y^2$
BP.PB' = $((b-x)\boldsymbol i + y\boldsymbol j).((-b-x)\boldsymbol i+y\boldsymbol j)$
BP.PB'= $x^2-b^2+y^2$
The second line does not follow from the first line and the fourth does not follow from the third.
Write out the intermediate steps between the first and second lines and you will see what went wrong.

Thanks for your reply andrewkirk. Unfortunately I'm still not able to identify my error. Here are the steps I omitted:

AP.PA' = $((a-x)\boldsymbol i + y\boldsymbol j).((-a-x)\boldsymbol i+y\boldsymbol j)$

The dot product is distributive over vector addition, so

AP.PA' = $(a-x)\boldsymbol i.(-a-x)\boldsymbol i+ y\boldsymbol j.(-a-x)\boldsymbol i+(a-x)\boldsymbol i.y\boldsymbol j+y\boldsymbol j.y\boldsymbol j$

$\boldsymbol i.\boldsymbol i=\boldsymbol j.\boldsymbol j=1$ and $\boldsymbol i.\boldsymbol j= 0$

so
AP.PA' = $x^2-a^2+y^2$

Alternatively this shorthand is suggested by my book;

if
$\boldsymbol a = x_1\boldsymbol i+ y_1\boldsymbol j$

and
$\boldsymbol b = x_2\boldsymbol i+ y_2\boldsymbol j$

then
$\boldsymbol a.\boldsymbol b=x_1x_2+y_1y_2$

so
$((a-x)\boldsymbol i + y\boldsymbol j).((-a-x)\boldsymbol i+y\boldsymbol j) =(a-x)(-a-x)+y^2$

$= -a^2+x^2+y^2$

andrewkirk