Prove Midpoint of Hyperbola Chord PQ is on Curve c2(x2+y2)+axy(a-2x)=0

[Michael_Light]

Homework Statement

P(cp , c/q) and Q(cq , c/q) are two points on the curve xy=c². Prove that the chord PQ has an equation pqy+x=c(p+q). A variable chord of the hyperbola xy=c² subtends a right angle at the fixed point (a,0). Show that the midpoint of the chord lies on the curve c²(x²+y²)+axy(a-2x)=0.

Homework Equations

The Attempt at a Solution

I managed to show that pqy+x=c(p+q) but failed to show that c²(x²+y²)+axy(a-2x)=0. I tried by by substituting the midpoint of points P and Q into c²(x²+y²)+axy(a-2x) but that leads to a rather complicated equation for me... another thing i can get is that ((c/p)/(cp-a))((c/q)/(cq-a))=-1 but yet, it doesn't seems to help me a lot in solving this question... can anyone give me some hints and some explanations? Thanks in advance..

 

[praharmitra]

What are doing is correct. Just work a little more. Show that substituting the midpoints of P and Q in [itex]c^2(x^2+y^2)+axy(a-2x)=0[/itex] gives you exactly the condition

[tex]
\frac{c/p}{cp-a}.\frac{c/q}{cq-a} = -1
[/tex]

 

[Michael_Light]

I still not managed to solve it... Can anyone help me...?

 

[praharmitra]

I still not managed to solve it... Can anyone help me...?

Just substitute [itex]x = \frac{c}{2}(p+q), y = \frac{c}{2}\left(\frac{1}{p} + \frac{1}{q}\right)[/itex] in [itex]c^2(x^2+y^2)+axy(a-2x)=0[/itex]. You should get
[tex]
\frac{c/p}{cp-a}.\frac{c/q}{cq-a} = -1
[/tex]

 

[Michael_Light]

Just substitute [itex]x = \frac{c}{2}(p+q), y = \frac{c}{2}\left(\frac{1}{p} + \frac{1}{q}\right)[/itex] in [itex]c^2(x^2+y^2)+axy(a-2x)=0[/itex]. You should get
[tex]
\frac{c/p}{cp-a}.\frac{c/q}{cq-a} = -1
[/tex]

Yea... I know... but by substituting [itex]x = \frac{c}{2}(p+q), y = \frac{c}{2}\left(\frac{1}{p} + \frac{1}{q}\right)[/itex] in [itex]c^2(x^2+y^2)+axy(a-2x)=0[/itex] , i get a very complicated equation which i failed to express it in the form [tex]\frac{c/p}{cp-a}.\frac{c/q}{cq-a} = -1[/tex]... Maybe i am missing something or there is some trick in solving this?