Conics Hyperbola Question (1 Viewer)

Joined
Nov 7, 2011
Messages
150
Location
New South Wales, Australia
Gender
Male
HSC
2013
I have no idea what I'm doing wrong with this question:

P( p, 1/p) and Q (q, 1/q) are two varibale points on the rectangular hyperbola xy = 1.
a) If M is the midpoint of the chord PQ and OM is perpendicular to PQ, express q in terms of p

I found:

M = ((p+q)/2, (p+q)/2pq)
m(OM) = 1/pq
m(PQ) also turns out to be 1/pq

Since the gradients are equal, they're parallel not perpendicular. What am I doing wrong?

Thanks in advance!
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
I have no idea what I'm doing wrong with this question:

P( p, 1/p) and Q (q, 1/q) are two varibale points on the rectangular hyperbola xy = 1.
a) If M is the midpoint of the chord PQ and OM is perpendicular to PQ, express q in terms of p

I found:

M = ((p+q)/2, (p+q)/2pq)
m(OM) = 1/pq
m(PQ) also turns out to be 1/pq

Since the gradients are equal, they're parallel not perpendicular. What am I doing wrong?

Thanks in advance!
Redo your working for m(PQ), it should turn out to be -1/pq

Which means that p=1/q (since -1/pq * 1/pq = -1)
 
Joined
Nov 7, 2011
Messages
150
Location
New South Wales, Australia
Gender
Male
HSC
2013
Redo your working for m(PQ), it should turn out to be -1/pq

Which means that p=1/q (since -1/pq * 1/pq = -1)
Oh my god! I found it like three times and all those times I completely forgot to keep the negative in I'm so stupid ugh.

Thanks again!

Also, I guess it would be +/- 1/q as opposed to 1/q?
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Oh my god! I found it like three times and all those times I completely forgot to keep the negative in I'm so stupid ugh.

Thanks again!

Also, I guess it would be +/- 1/q as opposed to 1/q?
Yep you are correct haha
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top