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First Principle Hyperbolas


Date: 01/05/99 at 08:58:55
From: Janis
Subject: Hyperbolas

Doctor Math,

I'd like to know the answer to this question. How do you solve for the 
standard equation of a hyperbola given its properties?

I know the standard equation already because we have discussed it at 
school. I just hope that your answer will come in a step-by-step 
process.
  
Thanks a lot in advance, 
Janis


Date: 01/05/99 at 10:27:56
From: Doctor Rob
Subject: Re: Hyperbolas

Thanks for writing to Ask Dr. Math!

I assume that by "its properties" you mean that a hyperbola is the 
locus of points such that the absolute difference between the distances 
from the two foci is a constant. In other words, if you take any point 
on the hyperbola, measure how far it is from each of the foci, and take 
the absolute difference of the two measurements, you will get the same 
constant. There are other definitions, and they lead to different 
derivations.

Let the foci F and G have coordinates (-c,0) and (c,0), where c > 0
(if c = 0, the foci coincide, and the locus is either the entire
xy-plane or empty). Let the general point P on the locus have 
coordinates (x,y). Let the constant be 2*a, where a >= 0 (you'll see 
that the 2 makes the equation come out nicer in the end). Then by 
considering the triangle PFG and using the Triangle Inequality,

   PF + FG >= PG, and PG + FG >= PF
   FG >= PG - PF, and FG >= PF - PG
   2*c = FG >= |PF-PG| = 2*a
   c >= a

Now use the distance formula to compute the distances PF and PG.
Then, by the above locus definition

   |sqrt([x+c]^2+y^2)-sqrt([x-c]^2+y^2)| = 2*a

Square both sides, use the fact that |Q|^2 = Q^2 for any real number Q, 
and expand

    [x+c]^2+y^2-2*sqrt([x+c]^2+y^2)*sqrt([x-c]^2+y^2)+[x-c]^2+y^2 
          = 4*a^2

Isolate the term with the square roots on the righthand side, gather 
like terms, and divide everything by 2:

   x^2 + c^2 + y^2 - 2*a^2 = sqrt([x+c]^2+y^2)*sqrt([x-c]^2+y^2)

Now square again, and all the square roots will go away:

   (x^2+c^2+y^2-2*a^2)^2 = ([x+c]^2+y^2)*([x-c]^2+y^2)

Now expand everything in sight, and move it all to the lefthand side, 
gathering like terms. Divide by 4, and you should have

   (c^2-a^2)*x^2 - a^2*y^2 - a^2*(c^2-a^2) = 0

Now recall that c >= a, so c^2 - a^2 >= 0. Put b = sqrt(c^2-a^2), so 
b >= 0, substitute c^2 - a^2 = b^2, and add a^2*b^2 to both sides to 
get

   b^2*x^2 - a^2*y^2 = a^2*b^2

At this point, everything we have done is fine. Now we want to divide 
by a^2*b^2, to get the locus equation

   x^2/a^2 - y^2/b^2 = 1

which is the standard form. This works if and only if both a and b are 
nonzero, if and only if c > a > 0. This is the answer you seek.

If a = 0, then b = c > 0, and we get the locus equation x^2 = 0, which 
is two copies of the line x = 0; that is, two coincident lines.

If a > 0 and b = 0 (so c = a), then we get the locus equation y^2 = 0, 
which is two copies of the line y = 0, again two coincident lines.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Equations, Graphs, Translations

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