Locus of the Midpoint of a Chord

Date: 03/13/98 at 09:15:48
From: Chye-Keat Chew
Subject: Conic section

Dear sir, 

Here is a question on conic sections: a variable chord of hyperbola 
x^2/a^2  - y^2/b^2 = 1 passes through the fixed point (h,k). Show that 
the locus of the midpoint of the chord is a hyperbola, with center 
(h/2,k/2). 

I don't know how to start to solve it, but I got some sort of a 
similar question with answer (see below). I really don't how to merge 
its solution with the problem above, though.

Question: A straight line passing through the fixed point (h,k) meets 
the x- and y-axes at P and Q, respectively. Show that the locus of the 
midpoint of PQ is a rectangular hyperbola with center (h/2, k/2).

Solution:

Let PQ have equation:

     x/2a + y/2b = 1    [intercept form with P(2a,0), Q(0,2b)].

Then

     h/2a + k/2b = 1.

Let M(x,y) be the midpoint of PQ, thus we have x = a and y = b. 
Therefore, at M,

     h/2x + k/2y = 1,

i.e., the locus of M has equation:

     xy - kx/2 - hy/2 = 0

     (x - h/2)(y - k/2) = hk/4.

Referring to parallel axes through (h/2,k/2), the above equation 
becomes

     xy = hk/4,

and therefore represents a rectangular hyperbola with center at the 
point (h/2, k/2) referred to the original axes.

Date: 03/13/98 at 17:28:56
From: Doctor Anthony
Subject: Re: Conic section

Let the chord have equation

            y - k = m(x - h)

Let (p,q) be its midpoint.

Now to get m (slope of chord), we have the property that if m' is 
slope of diameter conjugate to diameter of slope m, then

              mm' = b^2/a^2.

But the diameter of slope m' will pass through the midpoint of the 
given chord, so m' = q/p.

So:
           m(q/p) = b^2/a^2

                m = (b^2/a^2)(p/q)

Now (p,q) satisfies the equation of the chord, so we have:

            q - k = m(p - h)

            q - k = (b^2/a^2)(p/q)(p - h)

         q^2 - qk = (b^2/a^2)(p^2 - ph)

 q^2 - qk + k^2/4 = (b^2/a^2)(p^2 - ph + h^2/4)
                      + k^2/4 - (b^2/a^2)h^2/4

      (q - k/2)^2 = (b^2/a^2)(p - h/2)^2 = k^2/4 - (b^2/a^2)h^2/4

   (q - k/2)^2   (p - h/2)^2       k^2     h^2
    ---------  -  ----------  = ------ - ------.
       b^2           a^2          4b^2    4a^2

So reverting to coordinates (x,y) for the locus of midpoint (p,q) and 
changing signs throughout:

  (x - h/2)^2   (y - k/2)^2   (b^2)(h^2) - (a^2)(k^2)
   ---------  - ----------  = ----------------------
      a^2           b^2           4(a^2)(b^2)

If we let the right hand side be represented by the letter R, the 
equation of the locus can be written:

   (x - h/2)^2     (y - k/2)^2
   ----------   -  -----------  =  1,
    (a^2)(R)        (b^2)(R)

which is the equation of a hyperbola with center (h/2,k/2).

-Doctor Anthony, The Math Forum
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