1. ## Hyperbolas ...

Hello.

I was just wondering for this question:

A variable normal to xy = c^2 meets the x-axis and y-axis at P, Q. Show that the locus of midpoint of PQ is
Do I:
1. Find equ. of normal.
2. Find P + Q.
3. Find midpoint.
4. Somehow manipulate midpoint coordinates to get locus.

I just wanted to make sure that it is the right method because it seems to me to be pretty messy since my P + Q are kinda yucky already so by the time I got to step 4...*shudders*

2. Hello, UltraGirl!

Your game plan is a good one.

I gave it a quick try and got into a mess at step 4, too.

I'll work on it and try to get back to you.

3. Ummm is the midpoint = $[\frac{ct^4-c}{2t^3}, \frac{c-ct^4}{2t}]$?

4. Originally Posted by UltraGirl
Ummm is the midpoint = $[\frac{ct^4-c}{2t^3}, \frac{c-ct^4}{2t}]$?
I got the midpoint to be:
$
[\frac{t^4-c^4}{2t^3}, \frac{c^4-t^4}{2tc^2}]
$

5. Ummm I've reworked a couple of times and I still get the same midpoint

I've tried using your midpoint to find the locus but it doesnt seem to work out to be the one in the question

6. This is my working out so far. Could someone please check to see if it's right?

7. Originally Posted by UltraGirl
This is my working out so far. Could someone please check to see if it's right?
You midpoint differs to mine only because i started with the point

8. Ummm so someone suggested that I simply substituted the midpoint into the locus provided in the question, proving LHS = RHS, and that would solve the problem BUT how on earth do you do that with such high powers???

9. To check whether the given problem is correct or not, select a numerical problem.
Let the given equation be xy = 4. Draw a normal at (1, 4).
Then find the mid point of PQ and substitute in the given expression. See whether it satisfies or not.