Another Conics question. (1 Viewer)

[McSo]

Question: P(cp,c/p) and Q(cq,c/q) lie on the rectangular hyperbola xy=c^2. The chord PQ subtends a right angle at another point R (cr,c/r) on the hyperbola. Show that the normal at R is parallel to PQ.

I'm guessing the basic thing to was to find the gradients and prove they're equal.
I have M_pq = -1/pq and M_R = -1/r^2...
I'm guessing that I need to show pq = r^2 (If my gradients are correct)
So I tried substituting x = 2rp - (r^2)y in x+pqy=c(p+q)
(The equation of PQ => x+pqy = c(p+q) and the equation of the
normal r => x+(r^2)y = 2rp)

I ended up with y(r^2 - pq) = 2rp - c(p+q).. but I don't really know what I'm doing..

 

[Riviet]

Show that the normal at R is parallel to PQ.

Are you sure it's the normal at R? Because I drew a diagram and it appears the normal at R would be perpendicular to PQ, instead of parallel.

Edit: I probably interpretted the question wrong.

 

[who_loves_maths]

Looks like the normal is fine from here Riviet.

 

[McSo]

Positive, its from question 8 from exercise 3.4 from Cambridge.

 

[who_loves_maths]

This is why "rough" sketches are dangerous in maths...

The reason you think the normal is perpendicular to PQ is most probably because you are drawing the three points P, Q, R on the same branch of the hyperbola Riviet.

This cannot happen, as you will see with a more accurate diagram. The angle subtended at R can get close to a right angle, but actually never is.

P, and Q lie necessarily on different branches of the hyperbola.


McSo, this problem may be done using geometric arguments as well as algebraic ones... but I'll leave it all to Riviet as I'm sure he is vehemently pondering and typing away at the question right now.

 

[McSo]

Hmm.. I just drew a diagram.. (I didn't think I'd need one haha) but.. I realised if PQ is perpendicular to the tanget at R and you're trying to prove that the normal at R is parallel to PQ, then could you just say, because of the cointerior angles, the normal R must be parallel to PQ ?

 

[who_loves_maths]

?
Umm... unfortunately no McSo, since the condition that the tangent be perpendicular to PQ is exactly what you are trying to find in the first place!


But of course once you have shown that, then a simple geometric argument following that will suffice towards the final proof.
(i.e. that if PQ is perpendicular to the tangent at R then this implies that the normal at R is parallel to PQ since the normal and the tangent themselves are out of phase by a right angle.)

 

[McSo]

Ah ok... Well I'm stuck then I guess :| I can't find any link between the gradients.

 

[Riviet]

Ah, I see now, my mistake certainly was drawing all three points on the same branch. Anyway, I should be off to bed, got a morning class tomorrow. This one's all yours who loves maths.

 

[who_loves_maths]

Spoiler

Algebraic Mthd:
Gradient of PQ: [c(1/p - 1/q)]/[c(p - q)] = -1/pq ___________ (1)
Gradient of tangent @ R = dy/dx = d(c/r)/d(cr) = d(1/r)/dr = -1/r²
i.e. Gradient of normal @ R = r² ___________ (2)

Relationship btw P, Q, R:
grad(PR).grad(QR) = -1 = (-1/pr).(-1/qr) = 1/pqr² ---> pqr² = -1
---> r² = -1/pq ---> (2) = (1)

Since gradients are equal, both lines make same angle with positive x-axis. ---> The lines are parallel. 

EDIT:

Spoiler

Note that:
r² = -1/pq ---> pq < 0
---> p > 0 when q < 0, or, p < 0 when q > 0
---> P and Q lie on different branches of hyperbola.