Consider $A\in M_n(\mathbb{R})$ a positive definite matrix and a matrix $B\in M_{n \times p}(\mathbb{R})$, with $n\geq p$ and $rank(B)=p$.
i) Show that $C=B^TAB$ is positive definite.
ii) Show that the equation $x^TAx=1$ is a hyperellipse in $\mathbb{R}^n$.
For i), $$ x\neq 0 \Rightarrow Bx\neq 0. $$
So $$ \langle B^tABx,x\rangle = \langle A(Bx),Bx \rangle > 0.$$
But for ii) I have no idea.
1) The following works for $A$ symmetric.
We can diagonalize $A$: $$A = Q^TDQ,$$ where $Q$ is unitary and $D$ is diagonal. Such a factorization exists since $A$ is symmetric positive definite and therefore normal. Then $$x^TAx = x^TQ^TDQx = (Qx)^TDQx.$$ Let $y = Qx$. Then you have $$ y^TDy = 1.$$
Working things out in components, you should see that this is precisely the equation for a hyperellipse.
2) When $A$ is not symmetric, the situation is messier, but still doable.
Define $B:= \frac{1}{2}(A+A^T)$. $B$ is symmetric positive definite, so by part 1), $B$ gives rise to a hyperellipse. I claim $B$ and $A$ define the same hyperellipse.
Indeed, $$ \frac{1}{2}x^TA^Tx = \frac{1}{2}x^TAx ,$$ which you can see by working things out component-wise.
Then $$x^TBx:= \frac{1}{2}(x^TAx+x^TA^Tx) = x^TAx.$$