Angle between two parabolas I'm a little confused about a problem that asks me to find the angle between the two parabolas $$y^2=2px-p^2$$ and  $$y^2=p^2-2px$$ at their intersection. I used implicit differentiation to find the slopes $$y'=\frac{-p}{y}$$ and $$y'=\frac{p}{y}$$ Using the formula for the angle between two lines $\tan\alpha=|\frac{m_2-m_1}{1+m_2m_1}|$ with some substitutions I ended up with  $\tan\alpha=|\frac{y}{x}|$. When I set the two equations equal to one another to find the point of intersection I ended up with $x=0$ and $y=±p$. 
I know these values lead to tangent being undefined, can I infer from this fact that the angle between these parabolas is a right angle? The solution says that they intersect at right angles, but I'm confused about using something that is undefined to formulate my answer, if it turns out to be correct. Thanks for any help.

Notice, we have $$y^2=2px -p^2$$$$\implies 2y\frac{dy}{dx}=2p \iff \frac{dy}{dx}=\frac{p}{y}=m_1$$
$$y^2=p^2-2px$$$$\implies 2y\frac{dy}{dx}=-2p \iff \frac{dy}{dx}=\frac{-p}{y}=m_2$$ Now, the angle between the parabolas is given as $$\tan \alpha=\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$ $$=\left|\frac{\frac{p}{y}-\left(\frac{-p}{y}\right)}{1+\frac{p}{y}\left(\frac{-p}{y}\right)}\right|$$ $$=\left|\frac{2py}{y^2-p^2}\right|$$
On solving the equations of the parabolas we get intersection point $\left(\frac{p}{2}, 0\right)$  which is lying on the x-axis. 
Now, substituting the value of $y$ we get the angle between the curve at the point of intersection $$\tan \alpha=\left|\frac{2p(0)}{(0)^2-p^2}\right|$$  $$=0 \iff \alpha=\tan^{-1}(0)=0$$ Thus, the angle between the given parabolas at the point of intersection $\left(\frac{p}{2}, 0\right)$ is $\alpha=0$
$$\bbox[5px, border: 2px solid #C0A000]{\color{red}{\alpha=0}}$$
In other words, we can also say that the given parabolas: $y^2=2px-p^2$ & $y^2=p^2-2px$ are touching each other at the point $\left(\frac{p}{2}, 0\right)$.
