Use a horizontal normal to identify a vertical tangent

$dx/dt=-k\sin t$ and $dy/dt=2k\cos2t$. For $p\ne0$, the tangent slope is $-2\cos2p/\sin p$, so the normal slope is $\sin p/(2\cos2p)$. Since $-\pi/4<p<\pi/4$, $\cos2p>0$ and the normal equation is $y-k\sin2p=[\sin p/(2\cos2p)](x-k\cos p)$. This remains valid at $p=0$: the tangent velocity is $(0,2k)$, so its perpendicular normal is horizontal.

At $p=0$, the point is $(k,0)$ and the normal is $y=0$. The perpendicular tangent is therefore $x=k$. This conclusion uses the normal or the velocity; it does not assign a finite value to $dy/dx$ at zero.