Answer to Question #105292 in Analytic Geometry for Emmanuel

Question

Prove that the equation of the chord joining the points P(ct, c/t) and Q(cT, c/T) on
the rectangular hyperbola xy = c
2
is x + tT y = c(t + T). M is the midpoint of P Q
and P Q meets the x-axis at N. Prove that OM = MN, where O is the origin

Answer 1

P (ct,c/t) and Q (cT,c/T)

Slope of line joining PQ is $\dfrac{(c/T – c/t)}{(cT – ct)}= – 1/Tt$

 Equation of the line using slope-point form is $(y – ct) = (– 1/cTt)(x – c/t)\\$

Simplifying we get $x + tTy = c(t + T).$

M is the midpoint of PQ. By midpoint formula,

$M \equiv (c(t + T)/2 , c(1/t + 1/T)/2)$

i.e.$M(c(t + T)/2 , c(t + T)/2tT)$

By distance formula,

$OM = \sqrt{({\dfrac{{c(t + T)}}2 – 0})^2 + ({\dfrac{c(t + T)}{2tT} – 0})^2}$

Simplify to get OM = $[c(t + T)\sqrt{(1 + T^2t^2)] / 2}] …...(1)$

Put y = 0 in the equation of PQ for the coordinates of N.

$N \equiv (c(t + T),0)$

Similarly MN =$\sqrt{{(\dfrac{c(t + T)}2 – c(t + T)})^2 + {{(\dfrac{c(t + T)}{2tT} – 0}})^2}$

Simplifying we get MN $= [c(t + T)\sqrt{(1 + T^2t^2) / 2}] …...(2)$

From (1) & (2),

OM = MN