Prove that the midpoints of chords of the hyperbola (square of x)/(square of a)–(square of y)/(square of b) = 1 parallel to the diameter y = mx lie on the diameter (square of a)my = (square of b) x?

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Let the cord parallel to y=mx be y=mx+c..............(1) Then the points of intersection with the hyperbola are given by:- x^2/a^2 - (mx+c)^2/b^2 = 1 or (1/a^2-m^2/b^2)x^2-2mxc/b^2-c^2/b^2-1=0 .....(2) Let (x1,y1) and (x2,y2) be the end points of the cord. Then x1 and x2 both satisfy...
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Let the cord parallel to y=mx be y=mx+c..............(1) Then the points of intersection with the hyperbola are given by:- x^2/a^2 - (mx+c)^2/b^2 = 1 or (1/a^2-m^2/b^2)x^2-2mxc/b^2-c^2/b^2-1=0 .....(2) Let (x1,y1) and (x2,y2) be the end points of the cord. Then x1 and x2 both satisfy Eq.(2) So x1+x2 = [2mc/b^2]/(1/a^2-m^2/b^2) = 2mc/(b^2/a^2-m^2)...............(3) Let (h,k) be the midpoint of the cord. Then h=(x1+x2)/2 = mc/(b^2/a^2-m^2) and k=(y1+y2)/2. Now both (x1,y1) and (x2,y2) satisfy Eq(1) so that k=mh+c. Substituting the value of c from Eq(3) k=mh+(b^2/a^2-m^2)h/m = b^2/(a^2m) Or a^2 mk = b^2h Therefore the locus of the midpoint (h,k) of the cord is a^2 my = b^2x. read less
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