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.