Prove that the set $\{x : x^T(A-B)x = 1\}$ is not bounded 
Suppose we have two symmetric, positive definite $n \times n$ matrices $A$ and $B$, where $n > 1 $ and $A \neq B$. Also $\det A = \det B$. Prove that the set $$\{x : x^T(A-B)x = 1\}$$ isn't bounded. 

My approach was rather intuitive: $x^T(A-B)x =1 \rightarrow x^TAx - x^TBx = 1 \rightarrow (A^{\frac{1}{2}}x)^T(A^{\frac{1}{2}}x) - (B^{\frac{1}{2}}x)^T(B^{\frac{1}{2}}x) = 1 \rightarrow ||A^{\frac{1}{2}}x||^2 - ||B^{\frac{1}{2}}x||^2 = 1$ 
and this expression is similar to equation $x^2 - y^2 = 1$ which define a hyperbola which isn't bounded.
I understand that the last conclusion seems strange and extremely not rigorous, that's why I ask you to help me prove it rigorously. 
P.S. If my intuitive approach works out here, can we say that the set $\{x: x^T(A+B)x = 1\}$ is bounded (because $x^2 + y^2 = 1$ is bounded )
Thank you very much in advance for any help!
$\mathbf{UPDATE:}$ The proof of the fact that $A-B\ $ is neither positive semidefinite nor negative definite is here
 A: Your intuition is interesting, and you almost got it right. But there is a problem. 
If $T$ is symmetric positive definite, then the set $X=\{x:\ x^TTx=1\}$ is bounded. This can be seen by writing $T=W^TDW$, with $W$ orthogonal and $D$ diagonal. Then $Y=WX$, where $Y=\{y:\ y^TDy=1\}$. Thus $X$ is bounded if and only if $Y$ is bounded. And $Y$ consists of those $y$ such that $$\tag1d_1y_1^2+d_2y_2^2+\cdots+d_ny_n^2=1,$$ which is bounded. 
In particular, as you say, the case with $A+B$ produces a bounded set. 
The problem with $A-B$ arises since $A-B$ can still be positive definite. For instance, if $A=2B$, then your set is bounded. More generally, any choice such that $A-B$ is positive definite will produce a bounded set. 
When $A-B$ is not positive semidefinite nor negative definite, it is still symmetric so diagonalizable, and it will have a positive and an negative eigenvalue; thus one can reproduce the argument above but now in $(1)$ some terms will have negative signs, which will allow you to make the set unbounded. 
A: The statement given is false: take $A$ and $B$ to be equal $1 \times 1$ matrices containing the number "1" as their sole entry. Then there are no numbers with $x^t (A-B)x = 1$ (i.e., with $0x^2 = 1$), so the set you've got is empty, hence bounded. 
