another coordinate systems

Can someone show me how to do this question please. :confused:

Show that the chord joining P(ct1,c/t1) and Q(ct2,c/t2), where P and Q are on the curve with equation xy=c² has equation:

x+t1t2y=c(t1+t2)

Given that t1 and t2 vary but that PQ always passes through the point (2c,2c), show that the locus of the mid-point of PQ is the curve with equation xy=c(x+y).

Thanks

Reply 1

evariste_3086

dhokes

Can someone show me how to do this question please. :confused:

xy=c^2
let P(ct,c/t) Q(ct'.c/t')

grad PQ=c/t'-c/t/ct'-ct
=t-t'/tt'(t'-t)
=-1/t't

eqn of PQ=

y-c/t=-1/tt'[x-ct]
tt'y-ct'=-x+ct
tt'y+x=c(t+t') ........(*)

midpoint of PQ= [c(t+t')/2 ,c(t+t')/2tt']
now as t varies Pq always passes (2c,2c)
using (*) with x=2c y=2c
we get
2+2tt'=(t+t')............(1)
midpoint of PQ has co-ords
x=c(t+t')/2
y=c(t+t')/2tt'

using (1) on the x,y value

x=c(1+tt')
y=c(1+tt'/t't)

and hence
xy=c^2[1+2tt'+(tt')^2/tt']
=c^2[1/tt'+2+tt']
=c^2[1+tt'+1+1/t't]
=c[c(1+tt')+c(tt'+1)/tt']
=c(x+y)