prove that hyperbolic cone is affine set (check my solution)

Hyperbolic cone $C_P$ with $P$ positive definite matrix is a set that satisfies the following $C_P = \{x: x^TPx\leq (c^Tx)^2\}$. I need to prove that this set is affine. I know that this set is convex, because it is inverse image of second order cone under affine function $f(x)=(P^{1/2}x, c^Tx)$, However I couldn't show that this set is affine using the same technique.

I tried proving using definition of affinity. If $x_1,x_2 \in C_P$ then I need to show that $\theta x_1+(1-\theta)x_2$ should also be in $C_P$.

If $x_1,x_2 \in C_P$ then $2\theta x_1$ and $2(1-\theta)x_2$ should also be in $C_P$, it can be verified directly from definition of $C_P$. Now, because we know that the set is convex we can take combination of these two points with $\theta'=1/2$ and we will see that $\theta x_1+(1-\theta)x_2$ is in $C_P$. Is my solution correct?

Solution 1:

The set is not convex unless you add the condition $c^Tx \geq 0$. For example, for $P=I$ and $c=[\sqrt{2};0]$, $C_P = \{ x : x_2^2 \leq x_1^2\}$. The elements $(-1,1)$ and $(1,1)$ are in $C_P$, but $(0,1)$ is not.

The condition $c^Tx \geq 0$ ruins the affinity property.