Consider the following scalar term $$ T(x):=x^\top Qx, $$ with $x\in\mathbb{R}^2$, $x\neq 0$ and $$ Q:=\begin{bmatrix}-b\\a \end{bmatrix} \begin{bmatrix}-b+k_1&a+k_2 \end{bmatrix} \in\mathbb{R}^{2\times 2},$$
with fixed $a,b\in\mathbb{R}_{\geq 0}$, free $k_1,k_2\in\mathbb{R}$.
If $k_1=k_2=0$, it can be seen that $\mathrm{rank}(Q)=1$, since it corresponds to the outer product of the vector $\begin{bmatrix}-b\\a\end{bmatrix}$. Hence $Q$ has a positive and a zero eigenvalue, which implies that $T(x)$ is not (strictly) positive definite.
The question is now the following: does exist a pair $(k_1,k_2)$ (it can be dependent from $a,b$ if necessary) such that the term $T(x)$ is (strictly) positive definite? If not, can anyone provide a proof for non-existence?
EDITED
I've been proceeding by splitting the $Q$ in the symmetric and skew-symmetric parts $Q=Q_{sym}+Q_{skew}$ \begin{align} Q_{sym}:=&\begin{bmatrix}b^2-bk_1&-ab+\frac{ak_1-bk_2}{2}\\ -ab+\frac{ak_1-bk_2}{2}&a^2+ak_2 \end{bmatrix}\\ Q_{skew}:=&\begin{bmatrix}0&-\frac{ak_1+bk_2}{2}\\ \frac{ak_1+bk_2}{2}&0\end{bmatrix} \end{align}
Now, by using the fact that $x^\top Q_{skew}x=0$, the term reduces to $$ T(x)=x^\top Q_{sym}x, $$ where $Q_{sym}$ has rank 2.
It cannot be strictly positive definite.
$Q$ is the product of two matrices that are at most rank $1$, hence the rank of $Q$ has to be less than or equal to $1$.
Note that if $Q=AB$ then $rank(Q) \leq \min(rank(A),rank(B)) \leq 1$, hence $nullity(Q) \geq 1$.
We can find $x \neq 0$ such that $Qx=0$ and hence $x^TQx=0$.
Edit:
A simpler proof:
if $a=b=0$, clearly $Q=0$.
Otherwise, choose $x = \begin{bmatrix} a \\ b\end{bmatrix}$