conics question- locus problem (1 Viewer)

freaking_out

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alright, just bcoz i posted a tip in the "conics tip" section, it doesn't mean that i am good at conics.:mad1:

any way here is my question:

i) Prove that the locus of the midpoints of parallel chords of the rectangular hyperbola xy=c^2 is a diameter.

oh yeah, and a diameter means a straight line that passes through the centre of the hyperbola.
 

Affinity

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xy=c^2

let the set of parallel chords have common gradient m.
For a secant (chord extended) with Y-intercept b

the equation of the line is y= mx+b

this secant cut the rectangular hyperbolar at the solutions to the system of 2 equations, which also represents the extremities of the chord:

xy=c^2 ...............................(1)
y=mx +b ...............................(2)

x(mx+b) = c^2

mx^2 + bx - c^2 = 0

sum of roots = -b/m

whenever there is a pair of REAL solution
the mid point of the 2 extremities has x coordinate (-b/2m)

and hence y coordinate b/2 by substituting into (2)
hence the locus of the midpoints has parametric representation:

x = -b/(2m)
y = b/2
with b as the parameter.

eliminating b,
y = -mx

is a line which passess through (0,0)
QED.

well, not strictly a diameter in cases where m <0 because it doesn't intersect the hyperbola.
 

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