Rectangle hyperbola help

[Light_Bryant]

I got this problem and got stuck on part b.
Question: The point P (2p, 2/p) lies on rectangular hyperbola C with equation xy=4
a) find the equation of the normal to C at P.
I got p³x-py-2p⁴+2=0

The normal at P meets C again at the point Q. The mid point of PQ is the point M.
b) find in cartesian form, an equation of the locus of M as P varies.
Just need some help not the whole solution, thank you for taking your time to help me, greatly appreciated.

 

[Light_Bryant]

So after a while im closer to cracking it.
1) I re-arranged xy=4 to get x=4/y
2) substituted into p³x-py-2p⁴+2=0
3) got the points (2p, 2/p) and (-2p³,-2p³)
4) found the mid point ((-2p³+2p)/2 , (-2p³+2/p)/2)) now stuck lol

 

[stapel]

Question: The point P (2p, 2/p) lies on rectangular hyperbola C with equation xy=4
a) find the equation of the normal to C at P.
I got -p²x+y-2p³-2/p=0

I'm getting something slightly different. What were your steps?

 

[Light_Bryant]

I'm getting something slightly different. What were your steps?

Sorry I think part a was wrong here it is:
y=4/x
y=4x^-1
dy/dx=-4x^-2
-4/(2p)²
Gradient is -1/p²
Normal gradient is p²
y-2/p=p²(x-2p)
py-2=p³(x-2p)
py-2=p³x-2p⁴
p³x-py-2p⁴+2=0

 

[stapel]

I think part a was wrong

Yes.

y-2/p=p²(x-2p)

That's what I got, too. Note: Usually these types of equations are meant to be simplified to "y=" form. Are you sure you're supposed to get it into "0=" form? (If that's how your book or instructor is doing it, then the answer is "yes", and you should continue to do so. Otherwise, it's just extra work that you're safe not doing.)

Are you having trouble with part (b)? If so, kindly please reply showing what you've done with that. Thank you!

 

[DrPhil]

Sorry I think part a was wrong here it is:
y=4/x
y=4x^-1
dy/dx=-4x^-2
-4/(2p)²
Gradient is -1/p²
Normal gradient is p²
y-2/p=p²(x-2p)....OK to here
py-2=p³(x-2p)
py-2=p³x-2p⁴
p³x-py-2p⁴+2=0

Lets write the linear equation for the normal at point (2p,2/p):
$\displaystyle \displaystyle y_N = 2\left(\dfrac 1p - p^3\right) + p^2\ x$

This line intersects C at point Q (as well as at point P).

$\displaystyle \displaystyle y_Q = 2\left(\dfrac 1p - p^3\right) + p^2\ x_Q = \dfrac{4}{x_Q}$

Next step would be to solve that quadratic for $\displaystyle x_Q$. One or the roots is $\displaystyle x=2p$, so you might be able to divide out $\displaystyle (x - 2p)$.

 

[Light_Bryant]

Aha, never mind guy's managed to crack this problem by reading a maths book which was much more in-depth than the standard ones we have now.