Prove that the midpoint of PQ lies on the straight line $y+r^2x=0$.

Question

Prove that the equation of the chord joining the points $P(cp,\displaystyle\frac{c}{p})$ and $Q(cq,\displaystyle\frac{c}{q})$ on the rectangular hyperbola $xy=c^2$ is $pqy+x=c(p+q)$.

So, I've found that the gradient of the chord $PQ$ is $\displaystyle\frac{-1}{pq}$ and managed to prove that the equation of the chord is $pqy+x=c(p+q)$.

It is given that $PQ$ subtends a right angle to the tangent at the point $R(cr,\displaystyle\frac{c}{r})$ on the curve, prove that the midpoint of PQ lies on the straight line $y+r^2x=0$.

I've obtained that the midpoint of $PQ$ is $\displaystyle\big(\frac{cp+cq}{2},\frac{\frac{c}{p}+\frac{c}{q}}{2}\Big)$

Substitute $y=\displaystyle\frac{\frac{c}{p}+\frac{c}{q}}{2}$ into $y+r^2x=0$,

$\displaystyle\frac{\frac{c}{p}+\frac{c}{q}}{2}+r^2x=0$

$cp+cq=-2pqr^2x$

By substituting value of y into the equation, I expected to obtain the value of x, but how do I get rid of $r$?

How do I proceed? And how should I use the information given about the tangent at point R?

score 1 · Accepted Answer · answered Apr 01 '20 at 10:39

1

The gradient of $PR$ $$=\dfrac{c/p-c/r}{cp-cr}=-\dfrac1{pr}$$

Similarly, the gradient of $QR$ $$=-\dfrac1{qr}$$

So, we have $\dfrac1{pqr^2}=-1\iff pq=?$

If $M(h,k)$ is the midpoint $PQ,$

$$2h=c(p+q)$$

$$2k=\dfrac cp+\dfrac cq=?$$

$$\dfrac{2h}{2k}=\cdots=pq=-\dfrac1{r^2}$$

answered Apr 01 '20 at 10:39

lab bhattacharjee

274,582

Are you saying that PR is perpendicular to QR? – gc3941d Apr 01 '20 at 12:18
@gc3941d, I thought so – lab bhattacharjee Apr 01 '20 at 12:19
So the phrase in the question ‘PQ subtends a right angle to the tangent at the point R’ implies that PR is perpendicular to QR? – gc3941d Apr 01 '20 at 12:20
If it’s not too much to ask, how would PQ and R be drawn on a hyperbola? – gc3941d Apr 02 '20 at 02:28

Prove that the midpoint of PQ lies on the straight line $y+r^2x=0$.

1 Answers1