# Isosceles hyperbola locus problem

1. Nov 6, 2012

### dpesios

Hello all,

I would like beforehand to inform you that the translation of the following geometric problem is not very good and consequently you will have to use your mathematical intuition just a little bit. I encountered it while giving admission exams in a Mathematics department to pursue a second degree.
I don't have a mathematical background so any help appreciated.

1. The problem statement, all variables and given/known data
"Find the locus of points from which the conducted tangents at a given isosceles hyperbola are vertical to each other"

2. Relevant equations

3. The attempt at a solution
Can you assume that the locus contains just one point (which is the origin of the axes) and the tangents touch the curve at infinity ? :grumpy:
I apologize for my English ...

Last edited: Nov 6, 2012
2. Nov 6, 2012

### haruspex

All makes sense to me except for "isosceles hyperbola". Could it mean rectangular?
For the rest, I would word it as "Find the locus of points from which the subtended tangents to a given (?) hyperbola are perpendicular to each other"
No.

3. Nov 7, 2012

### dpesios

Yes, I do mean rectangular hyperbola.
When it comes to geometry it is all Greek to me ...

I'm not sure whether we should say "conducted tangent from a given point" or "subtended tangent".

As far as I understand two tangent lines start from a point on the locus and touch the given curve in a way that they are perpendicular to each other.

Any suggestions ?

4. Nov 7, 2012

### haruspex

Let (x,y) be a point on the locus, (x',y'), (x", y") be the contact points on the hyperbola of the conducted/subtended tangents. We have the following equations (assuming a canonical form for the hyperbola):
y'2-x'2=c2
y"2-x"2=c2
y'y - x'x = c2
y"y - x"x = c2
x'x"+y'y" = 0 (tangents perpendicular)
In principle, should be possible to eliminate x', x", y', y" to obtain an equation for x and y.

5. Nov 9, 2012

### dpesios

I see your point but there is always a but ...

No matter how hard I try, I cannot figure out an equation by eliminating x',x'',y' and y''.

Maybe there is another way of solving the problem.

I don't know. What do you think ?

6. Nov 9, 2012

### haruspex

I worked it through and got x=y=0. Then I thought about it geometrically and realised that is right. The gradient of the curve y2=x2+c2 has magnitude <= 1 everywhere. So two tangents at right angles must have gradients +1 and -1. In the more general hyperbola y2=m2x2+c2, only if m>1 does the locus become more interesting.

7. Nov 10, 2012

### dpesios

So, the answer is that there is no locus or the locus contains just one point and the contact points with the curve are at infinity as I initially proposed (?).

If it is so, I must play lottery.