Let us work out the equation of the intersection of a cone and plane in terms of some plane coordinates $(u,v)$ and see if it is a parabola indeed. Place a coordinate system on the apex of the cone.
Place the axis of the cone along the $\hat{j}$ direction, and slice it with a plane rotated an angle half of the cone angle $\psi$
The plane exists at a fixed distance $d$ from the origin, along the direction
$$ \hat{n} = \pmatrix{ \text{-}\cos \frac{\psi}{2} \\ \;\sin \frac{\psi}{2} \\ 0} $$
This defines the out of plane direction and also the two in-plane directions of $$ \hat{e} = \pmatrix{ \sin \frac{\psi}{2} \\ \cos \frac{\psi}{2} \\ 0} $$ and $\hat{k}$ which is common between the global coordinates and the plane coordinates.
We define a point in space $\boldsymbol{r}$ to be on the plane if it expressed as
$$ \boldsymbol{r} = d\,\hat{n} + u\,\hat{k} + v\,\hat{e} \tag{1}$$
where $(u,v)$ are the in-plane coordinates of the point. You can prove the above as it obeys the vector plane equation of $\boldsymbol{r} \cdot \hat{n} = d$ given a plane unit normal vector $\hat{n}$ and distance $d$ from the origin ($\cdot$ is the vector dot product).
Now these points must also obey the equation of a cone, which in this case can be written as
$$ \boldsymbol{r} \cdot \hat{j} = \| \boldsymbol{r} \| \cos \tfrac{\psi}{2} \tag{2}$$
The above states that the angle between the vector $\boldsymbol{r}$ and $\hat{n}$ should equal to $\psi/2$ for all $\boldsymbol{r}$.
The we can use the plane points from (1) in the cone equation (2) to get
$$ \begin{gathered}\left(d\,\hat{n}+u\,\hat{k}+v\,\hat{n}\right)\cdot\hat{j}=\|d\,\hat{n}+u\,\hat{k}+v\,\hat{n}\|\,\cos\tfrac{\psi}{2}\\
d\,\left(\hat{n}\cdot\hat{j}\right)+u\,\left(\hat{k}\cdot\hat{j}\right)+v\,\left(\hat{n}\cdot\hat{j}\right)=\sqrt{d^{2}+u^{2}+v^{2}}\,\cos\tfrac{\psi}{2}\\
v\cos\tfrac{\psi}{2}+d\sin\tfrac{\psi}{2}=\sqrt{d^{2}+u^{2}+v^{2}}\,\cos\tfrac{\psi}{2}\\
v+d\,\tan\tfrac{\psi}{2}=\sqrt{d^{2}+u^{2}+v^{2}}\\
\left(v+d\,\tan\tfrac{\psi}{2}\right)^{2}=d^{2}+u^{2}+v^{2}\\
\left(d\,\tan\tfrac{\psi}{2}\right)^{2}+2\left(d\,\tan\tfrac{\psi}{2}\right)\,v=u^{2}+d^{2}\\
v=\frac{u^{2}+d^{2}-\left(d\,\tan\tfrac{\psi}{2}\right)^{2}}{2\left(d\,\tan\tfrac{\psi}{2}\right)}
\end{gathered}$$
which is clearly a parabola since it is of the form $v = a u^2 + b$ with coefficients $a = \tfrac{1}{2 d} \cot \tfrac{\psi}{2}$ and $b = d \cot \psi$ which are constants derived from the geometry of the problem.
