Derivation of rotated hyperbola from locus of points definition

I am trying to write the equation of the set of points that has a constant difference in distance from two points (a hyperbola). I would like a general expression, no matter if the line joining the two points is parallel to an axis.

Calling the fixed points A = (a_x, a_y) and B = (b_x, b_y), and a point on the hyperbola P = (p_x, p_y), and the constant distance difference 'c', I started with:

$\displaystyle \sqrt{(a_x - p_x)^2 + (a_y + p_y)^2} - \sqrt{(b_x - p_x)^2 + (b_y + p_y)^2} = c$

$\displaystyle \sqrt{(a_x - p_x)^2 + (a_y + p_y)^2} = \sqrt{(b_x - p_x)^2 + (b_y + p_y)^2} + c$

Squaring both sides:

$\displaystyle (a_x - p_x)^2 + (a_y + p_y)^2 = c^2 + 2c\sqrt{(b_x - p_x)^2 + (b_y + p_y)^2} + (b_x-p_x)^2 + (b_y-p_y)^2$

As you can see, this is quite a mess. I am trying to get it to a more sane "general conic" form like:

$\displaystyle Ax^2 + Bxy + Cy^2 + Dx + Ey + F=0$

but if I square the whole thing again to get rid of the square root on the right, I'll introduce ^4 on the left, which seems like a bad plan. Any clues on where to go from here?

Thanks,

David

Re: Derivation of rotated hyperbola from locus of points definition

DAVIDDORIA

The points A, and B are the foci of the hyperbola...it could be better if you chose their coordinates to be A(-a,0) and B(a,0).....

you may continue to square one more time ...dont warry the fourth powers will be canceled..at the end and you will end up with the equation of the hyperbola you are looking to find .

Good Luck

MINOAS

Re: Derivation of rotated hyperbola from locus of points definition

Minoas - so if I do a change of coordinates so that the foci are at (-a,0) and (a,0), I could use that transformation to convert single points back to the original coordinate system, but how would I develop an expression for the hyperbola in the original coordinate system?

There must be something better than trying to expand something like this:

$\displaystyle \left(c^2 + 2c\sqrt{(b_x - p_x)^2 + (b_y + p_y)^2} + (b_x-p_x)^2 + (b_y-p_y)^2\right)^2$

Not only would it be 10 pages long, it would still contain the square root I am trying to get rid of!

David

Re: Derivation of rotated hyperbola from locus of points definition

try to use a coordinate system so that the hyperbola's centre will be the origin and derive the simplest known formula for the hyperbola. Then change it BAC K to the original coordinate system to find a formula AX^2+2BXY+CY^2+DX+EY+F=0 which is the general quadratic formula for all conics.I did this in my book and it is ok...

Try it. do not forget that the transformations are linear and therefore the degree of the curve is preserved....

Minoas