Solve the given problem involving parametric equations

[chwala]

Homework Statement

$P,Q,R$ and $S$ are four points on the hyperbola $x=ct, y=\dfrac{c}{t}$ with parameters $p,q,r$ and $s$ respectively. Prove that, if the chord $PQ$ is perpendicular to the chord $RS$, then $pqrs=-1$.

Relevant Equations

parametric equations

My take;

$y=\dfrac{c^2}{x}$

$y+x\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}=\dfrac{-y}{x}$

$y-\dfrac{c}{t}=-\dfrac{y}{x}(x-ct)$

$yt-c=-\dfrac{yt}{x}(x-ct)$

$xyt-cx=-yt(x-ct)$

$c^2t-cx=-cx+yct^2$

$c^2t-cx=-cx+ytct$

$c^2t-cx=-cx+c^2t$

$⇒-cx=-cx$

$⇒cx=cx$

Therefore it follows that,

$\dfrac{c}{c}=\dfrac{x}{x}$

$x=c$

$y$ will be given by,

$y=\dfrac{c^2}{c}=c$

point P will have co-ordinates $(x,y)=(c,c)$ and point Q will have co-ordinates $(x,y)=(-c,-c)$ where gradient is given by;

$m=\dfrac{c--c}{c--c}=\dfrac{2c}{2c}=1$

It follows that the perpendicular to the chord RS will have gradient =-1.

I do not have the solution to this question...your input is highly appreciated...

 

[Steve4Physics]

Non-calculus solution:

The 4 points are: $P(cp, \frac cp), Q(cq, \frac cq), R(cr, \frac cr)$ and $S(cs, \frac cs)$.

Start by finding the gradient of $PQ$: $m_{PQ} = \frac { \Delta y}{\Delta x}$ where, for example, $\Delta x= cq - cp = c(q-p)$.

Take it from there !

 

[chwala]

Non-calculus solution:

The 4 points are: $P(cp, \frac cp), Q(cq, \frac cq), R(cr, \frac cr)$ and $S(cs, \frac cs)$.

Start by finding the gradient of $PQ$: $m_{PQ} = \frac { \Delta y}{\Delta x}$ where, for example, $\Delta x= cq - cp = c(q-p)$.

Take it from there !

Ok i am getting the following;

Gradient of chord $PQ=\left[\dfrac{-1}{qp}\right]$ now if $PQ$ is perpendicular to chord $RS$ then the product of their gradients =$-1$, therefore,Gradient of chord $RS=\left[\dfrac{-1}{sr}\right]=-1, ⇒sr=1$

...therefore $pqrs=-1$.

 

[Steve4Physics]

Gradient of chord $RS=\left[\dfrac{-1}{sr}\right]=-1, ⇒sr=1$

...therefore $pqrs=-1$.

Correct final conclusion but the method is wrong.

$RS=\left[\dfrac{-1}{sr}\right]=-1, ⇒sr=1$ doesn't work. For example the gradient of SR could be 3 and the gradient of PQ could be $-\frac 13$.

Remember, you are being asked to prove "that, if the chord PQ is perpendicular to the chord RS, then pqrs= -1".

 

[chwala]

I do not seem to get it...

Ok, let $m_1$ and $m_2$ be the gradients of $PQ$ and $RS$ respectively;

$m_1=\left[\dfrac{-1}{qp}\right]$

$m_2=\left[\dfrac{-1}{sr}\right]$

we know that $m_1×m_2=-1$ therefore,

$\dfrac{-1}{qp}×\dfrac{-1}{sr}=-1$

$\dfrac{1}{qp}×\dfrac{1}{sr}=-1$

$\dfrac{1}{pqsr}=-1$

on cross-multiplying we end up with;

$pqsr=-1$.

 

[Chestermiller]

The vector from P to Q is $(q-p)c\hat{i}+\left(\frac{1}{q}-\frac{1}{p}\right)c\hat{j}$. The vector from R to S is $(s-r)c\hat{i}+\left(\frac{1}{s}-\frac{1}{r}\right)c\hat{j}$. For these chords to be perpendicular, the dot product of these two vectors must be zero.