[proofplan]
We first use the factorization $Q=L^\top L$ to rewrite the quadratic form as the square of the Euclidean norm of the affine expression $A(x)=L(Fx+g)$. Then we introduce an auxiliary scalar $t$ that upper-bounds this norm and separately satisfies the scalar bound $t^2 \leq \rho(x)$. Finally, we verify both inclusions between the original feasible set and the projection of the lifted conic set, encode the scalar quadratic inequality and the nonnegativity of its right-hand side with a rotated second-order cone, and obtain the sign condition on $t$ from the standard Lorentz cone constraint.
[/proofplan]
[step:Rewrite the quadratic form as a squared Euclidean norm]
Define the affine map $y: \mathbb{R}^n \to \mathbb{R}^q$ by $y(x)=Fx+g$, define the affine map $A: \mathbb{R}^n \to \mathbb{R}^m$ by $A(x)=L(Fx+g)$, and define the affine scalar map $\rho: \mathbb{R}^n \to \mathbb{R}$ by $\rho(x)=h^\top x+r$. The assumption $Q=L^\top L$ is meaningful because the statement assumes such a matrix $L \in \mathbb{R}^{m \times q}$ has already been chosen; the proof uses this factorization and does not require constructing $L$.
Define the original feasible set $S \subset \mathbb{R}^n$ by
\begin{align*}
S = \{x \in \mathbb{R}^n : (Fx+g)^\top Q(Fx+g) \leq h^\top x+r \text{ and } h^\top x+r \geq 0\}.
\end{align*}
Define the lifted feasible set $T \subset \mathbb{R}^n \times \mathbb{R}$ by
\begin{align*}
T = \{(x,t) \in \mathbb{R}^n \times \mathbb{R} : |A(x)| \leq t, \ t^2 \leq \rho(x), \text{ and } t \geq 0\}.
\end{align*}
For each $x \in \mathbb{R}^n$, since $Q=L^\top L$, we have
\begin{align*}
y(x)^\top Q y(x) = y(x)^\top L^\top L y(x).
\end{align*}
By the definition of the Euclidean norm on $\mathbb{R}^m$,
\begin{align*}
y(x)^\top L^\top L y(x) = |L y(x)|^2 = |L(Fx+g)|^2 = |A(x)|^2.
\end{align*}
Therefore the original quadratic inequality is equivalent to
\begin{align*}
|A(x)|^2 \leq \rho(x).
\end{align*}
[/step]
[step:Lift every original feasible point by choosing the norm as the auxiliary variable]
Let $x \in S$. Define the scalar $t \in \mathbb{R}$ by
\begin{align*}
t = |A(x)|.
\end{align*}
Then $t \geq 0$ by nonnegativity of the Euclidean norm, and $|A(x)| \leq t$ holds with equality. Since $x \in S$, the previous step gives
\begin{align*}
t^2 = |A(x)|^2 = (Fx+g)^\top Q(Fx+g) \leq \rho(x).
\end{align*}
Thus $(x,t) \in T$. Hence every $x \in S$ belongs to the projection of $T$ onto the $x$-coordinates.
[/step]
[step:Project every lifted feasible point back to the original quadratic constraint]
Let $(x,t) \in T$. From the constraint $|A(x)| \leq t$ and the condition $t \geq 0$, squaring both sides preserves the inequality and gives
\begin{align*}
|A(x)|^2 \leq t^2.
\end{align*}
The defining inequality $t^2 \leq \rho(x)$ then implies
\begin{align*}
|A(x)|^2 \leq \rho(x).
\end{align*}
Using the norm identity from the first step, this is exactly
\begin{align*}
(Fx+g)^\top Q(Fx+g) \leq h^\top x+r.
\end{align*}
Also, since $t^2 \geq 0$ and $t^2 \leq \rho(x)$, we obtain $\rho(x) \geq 0$. Therefore $x \in S$.
[guided]
We now prove the reverse implication carefully, because this is the step where the auxiliary variable must not introduce extra projected points. Recall that $A: \mathbb{R}^n \to \mathbb{R}^m$ is the affine map defined by $A(x)=L(Fx+g)$ and that $\rho: \mathbb{R}^n \to \mathbb{R}$ is the affine scalar map defined by $\rho(x)=h^\top x+r$. Also recall that the lifted feasible set is
\begin{align*}
T = \{(x,t) \in \mathbb{R}^n \times \mathbb{R} : |A(x)| \leq t, \ t^2 \leq \rho(x), \text{ and } t \geq 0\}.
\end{align*}
Suppose $(x,t) \in T$. By definition of $T$, the scalar $t$ satisfies
\begin{align*}
|A(x)| \leq t.
\end{align*}
The same definition also includes $t \geq 0$. Since the Euclidean norm is nonnegative, both sides of $|A(x)| \leq t$ are nonnegative, so squaring is order-preserving. Hence
\begin{align*}
|A(x)|^2 \leq t^2.
\end{align*}
The lifted scalar inequality gives
\begin{align*}
t^2 \leq \rho(x).
\end{align*}
Combining the two inequalities yields
\begin{align*}
|A(x)|^2 \leq \rho(x).
\end{align*}
Now substitute the definition $A(x)=L(Fx+g)$ and use $Q=L^\top L$. This gives
\begin{align*}
|A(x)|^2 = |L(Fx+g)|^2 = (Fx+g)^\top L^\top L(Fx+g) = (Fx+g)^\top Q(Fx+g).
\end{align*}
Therefore
\begin{align*}
(Fx+g)^\top Q(Fx+g) \leq h^\top x+r.
\end{align*}
Finally, the original statement also requires the right-hand side to be nonnegative. This follows from $t^2 \leq \rho(x)$ because $t^2 \geq 0$, so
\begin{align*}
0 \leq t^2 \leq \rho(x).
\end{align*}
Thus $\rho(x)=h^\top x+r \geq 0$, and $x$ satisfies both original constraints.
[/guided]
[/step]
[step:Represent the scalar quadratic inequality by a rotated cone]
Define the rotated second-order cone in dimension three by
\begin{align*}
\mathcal{Q}_r^3 = \{(a,b,s) \in \mathbb{R} \times \mathbb{R} \times \mathbb{R} : a \geq 0, \ b \geq 0, \ 2ab \geq s^2\}.
\end{align*}
For a fixed pair $(x,t) \in \mathbb{R}^n \times \mathbb{R}$, set
\begin{align*}
a = \frac{\rho(x)}{2}, \qquad b = 1, \qquad s = t.
\end{align*}
Then $(a,b,s) \in \mathcal{Q}_r^3$ is equivalent to
\begin{align*}
\frac{\rho(x)}{2} \geq 0, \qquad 1 \geq 0, \qquad 2\frac{\rho(x)}{2} \geq t^2.
\end{align*}
This is exactly
\begin{align*}
\rho(x) \geq 0, \qquad t^2 \leq \rho(x).
\end{align*}
Thus the rotated cone represents the scalar quadratic bound together with the nonnegativity of its right-hand side. It does not impose $t \geq 0$, because the third coordinate $s$ in $\mathcal{Q}_r^3$ is unrestricted in sign. Define the standard Lorentz cone in dimension $m+1$ by
\begin{align*}
\mathcal{L}^{m+1} = \{(u,z) \in \mathbb{R} \times \mathbb{R}^m : |z| \leq u\}.
\end{align*}
The constraint $(t,A(x)) \in \mathcal{L}^{m+1}$ is exactly $|A(x)| \leq t$, and it forces $t \geq 0$ because $|A(x)| \geq 0$. Therefore the conic system consists of the Lorentz cone constraint $(t,A(x)) \in \mathcal{L}^{m+1}$ and the rotated cone constraint $(\rho(x)/2,1,t) \in \mathcal{Q}_r^3$. Combining the previous two steps shows that the projection of this conic system is precisely $S$.
[/step]
