Isosceles hyperbola locus problem

[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.

Homework Statement

"Find the locus of points from which the conducted tangents at a given isosceles hyperbola are vertical to each other"

Homework Equations

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 ?
I apologize for my English ...

 

[haruspex]

"Find the locus of points from which the conducted tangents at a given isosceles hyperbola are vertical to each other"

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"

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 ?

No.

 

[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 ?

 

[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.

 

[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 ?

 

[haruspex]

I worked it through and got x=y=0. Then I thought about it geometrically and realized 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.

 

[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.

Thank you for your time.