How to Tell That This is a Hyperbola?

Question

{${(x_1,x_2) \in \Re^2 : x_1x_2 = 1, x_1 >0}$}

I'm trying to prove that the set is closed but I have trouble visualizing these sets. My book says this "is the branch of the hyperbola that lies in the first quadrant." How can I tell this? Is it from $x_1x_2 =1$

Answer 1

hint $$x_1x_2=1$$ you can write it like $$ \underbrace {(x_1+x_2)^2-(x_1-x_2)^2}_{\color{blue}{=4x_1x_2}}=4$$ Substitute $x=x_1+x_2$ and $y=x_1-x_2$: $$x^2-y^2= 4$$ $$(\frac x2)^2-(\frac y2)^2=1$$ Clearly an hyperbola $$x'^2-y'^2= 1 $$ Since $x_1>0$ we have $x_2>0$ and the curve is in the first quadrant

Answer 2

The sign of the discriminant $B^2-4AC$ of the general conic $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ lets you determine the type of conic. In this case, it equals $1\gt0$, so the equation is that of a hyperbola.