Locus of the Midpoint of a ChordDate: 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 Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/