A spectrahedron is a quadric cone when matrices in LMI are in $\mathbb{R}^{2\times 2}$ I am watching a lecture (just at the beginning around 0:50-0:57).    
The note says when $n=2$, $\mathcal{S}$ is a quadric cone; however, it seems that the professor says "a quadratic cone".   
On page 31 of Boyd & Vandenberghe's textbook Convex Optimization, the quadratic cone is also called the second order cone. 
However, I try to multiply out the formula of second order cone, I cannot obtain the definition of spectrahedra.    
And I have no idea of understanding how to view a spectrahedron as a quadric cone when $n = 2$.
 A: Consider the linear matrix inequality (LMI)
$$\mathrm X \succeq 0$$
where $\mathrm X \in \mathbb R^{2 \times 2}$ is symmetric. As $\mathrm X$ is symmetric, we have $\binom{3}{2} = 3$, rather than $2^2 = 4$, degrees of freedom. Hence, we write $\mathrm X \succeq 0$ as follows
$$\begin{bmatrix} x_{11} & x_{12}\\ x_{12} & x_{22}\end{bmatrix} \succeq 0$$
If $\mathrm X$ is positive semidefinite, then its $3$ principal minors are nonnegative,
$$x_{11} \geq 0 \qquad \qquad x_{22} \geq 0 \qquad \qquad \det(\mathrm X) = x_{11} x_{22} - x_{12}^2 \geq 0$$
Since we have a quadratic inequality, we can call the following spectrahedron
$$\left\{ (x_{11}, x_{12}, x_{22}) \in \mathbb R^3 \,\,\bigg|\,\, x_{11} \begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix} + x_{12} \begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix} + x_{22} \begin{bmatrix} 0 & 0\\ 0 & 1\end{bmatrix} \succeq 0 \right\}$$
the quadratic positive semidefinite cone. Is it actually a cone? Let's plot its surface

It looks like a cone to me.
However, what Boyd & Vandenberghe call the second-order cone is a cone in $\mathbb R^{n+1}$ that can be encoded by the following linear matrix inequality (LMI) in $(x,t)$
$$\begin{bmatrix} t & x^T\\ x & t I_n\end{bmatrix} \succeq 0$$
which produces the inequalities
$$t \geq 0 \qquad \qquad \qquad t^2 - x^T x \geq 0$$
or, alternatively, $t \geq 0$ and $\|x\|_2 \leq t$.
