Boundedness of a set

Question

Let $$S = \{(x, y) \in \mathbb{R}^2: x^2 + 2hxy + y^2 =1\}$$

For what values of $h$ is the set $S$ nonempty and bounded?

For $h = 0,$ it is surely bounded, the curve being the unit circle. What for other $h$?

Please help someone.

You didn't specify whether $(x,y)\in\mathbb R²$ or $\in\mathbb C^2$ (or anything more exotic). — Hagen von Eitzen, Sep 20 '15 at 15:10

score 5 · Answer 1 · answered Sep 20 '15 at 15:10

5

Hint. You may rewrite $$ x^2 + 2hxy + y^2 =1 $$ as $$ (x+hy)^2 + (1-h^2)y^2 =1 $$ to see that the elliptic case comes when $$ 1-h^2>0. $$

answered Sep 20 '15 at 15:10

Olivier Oloa

can you elaborate please?I mean how do you say 'elliptic case comes when $1-h^2>0$'? – ramanujan Nov 11 '15 at 16:54
1

@ramanujan If $1-h^2>0$ then the equation is of the form $X^2+Y^2=R^2$ which resembles the equation of a circle/an ellipse, otherwise you have $X^2-Y^2=R^2$ which is rather the equation of an hyperbolic curve. – Olivier Oloa Nov 11 '15 at 17:01
@olivierOh! I got it...Thank you – ramanujan Nov 11 '15 at 17:07
@ramanujan You are welcome! – Olivier Oloa Nov 11 '15 at 17:08

score 2 · Answer 2 · answered Sep 20 '15 at 15:15

2

Hint:

$x^2+2hxy+y^2-1=0$ is a conic, and it is an ellipse if its discriminant $B^2-4AC=4(h^2-1)$ is negative.

answered Sep 20 '15 at 15:15

Emilio Novati

2 Answers2