# Hyperbola and dot product

• Appleton
In summary, the homework statement says that if the distances of a point P from the points A(a, 0), A'(-a, 0), B(b, 0) B'(-b, 0) are related by the equation AP.PA'=BP.PB'. The locus of P is a hyperbola and the equations of its asymptotes are y =+-x and x =+-y.

## 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:
Appleton said:
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
$((a-x)\boldsymbol i + y\boldsymbol j).((-a-x)\boldsymbol i+y\boldsymbol j) =(a-x)(-a-x)+y^2$

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

Actually you're quite correct. My mistake. I can't see any error in your calculations.

However, I have another idea. The question refers to the distances of P from the four points, not to the displacement vectors. If they are being careful with their words then the items AP, PA', BP, PB' are to be interpreted as scalar amounts, not vectors, and the dot between them is to be interpreted as simple multiplication (not a dot product).

I suggest trying what happens when you make that interpretation.

Thanks for that, it makes sense now, so the asymptotes must be y =+-x

## 1. What is a hyperbola?

A hyperbola is a type of mathematical curve that is defined by the equation x2/a2 - y2/b2 = 1. It has two branches that are symmetrical about the x-axis and y-axis.

## 2. How is dot product used in hyperbolas?

In hyperbolas, the dot product is used to determine the angle between two vectors that intersect at a point on the hyperbola. It can also be used to find the equation of the tangent line to a hyperbola at a given point.

## 3. What is the difference between a hyperbola and an ellipse?

A hyperbola has two branches that are open and never intersect, while an ellipse has a closed curve that forms a loop. The equation for a hyperbola is also different from that of an ellipse.

## 4. Can a hyperbola have a negative value?

Yes, a hyperbola can have negative values for its x and y coordinates, as well as for its asymptotes. The branches of a hyperbola can also extend infinitely in both directions, including in the negative direction.

## 5. How is a hyperbola used in real life?

Hyperbolas have many real-life applications, such as in physics, engineering, economics, and astronomy. They can be used to model the orbit of planets, the trajectory of projectiles, and the growth of bacterial populations. They are also used in designing structures such as bridges and arches.