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Prove that the equation of the chord joining the points P(ct,c/t) and Q(cT,c/T) on the rectangular hyperbola xy = c2 is x + tTy = c(t + T). M is the midpoint of PQ and PQ meets the x-axis at N. Prove that OM = MN, where O is the origin

Prove that the equation of the chord joining the points P(ct,c/t) and Q(cT,c/T) on the rectangular hyperbola xy = c2 is x + tTy = c(t + T). M is the midpoint of PQ and PQ meets the x-axis at N. Prove that OM = MN, where O is the origin

Grade:12th pass

1 Answers

Samyak Jain
333 Points
5 years ago
P (ct,c/t) and Q (cT,c/T). 
Slope of line joining PQ is (c/T – c/t)/(cT – ct) = (t – T)/{(Tt)(T – t)} = – 1/Tt
\therefore 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[{c(t + T)/2 – 0}2 + {c(t + T)/2tT – 0}2]
Simplify to get OM = [c(t + T)\sqrt(1 + T2t2)] / 2        …...(1)
Put y = 0 in the equation of PQ for the coordinates of N.
N \equiv (c(t + T),0)
Similarly MN = \sqrt{}[{c(t + T)/2 – c(t + T)}2 + {c(t + T)/2tT – 0}2]
Simplifying we get MN = [c(t + T)\sqrt(1 + T2t2)] / 2        …...(2)
From (1) & (2),
OM  =  MN .
Pls approve.

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