Answer to Question #105290 in 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,

$M≡(c(t+T)/2,c(1/t+1/T)/2)$

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

By distance formula,

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

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.

$N≡(c(t+T),0)$

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

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

From (1) & (2),

OM = MN