Derive an ellipse tangent without losing its vertical cases

$\cos t=x/2$ and $\sin t=y$, so $x^2/4+y^2=1$. The half-open parameter interval traces the complete ellipse once; $(2,0)$ is included at $t=0$.

The ellipse is centred at the origin, with intercepts $(\pm2,0)$ and $(0,\pm1)$. Increasing $t$ traces it counterclockwise from $(2,0)$.

For $\sin t\ne0$, the derivative is $dy/dx=-\cos t/(2\sin t)$. At $(2\cos t,\sin t)$, multiply $y-\sin t=-[\cos t/(2\sin t)](x-2\cos t)$ to obtain $(\cos t)x+(2\sin t)y=2(\cos^2t+\sin^2t)=2$. At $t=0,\pi$, the slope division is invalid but the velocity is vertical and the same formula gives $x=2$ and $x=-2$, respectively.

At $P$, $\cos(7\pi/6)=-\sqrt3/2$ and $2\sin(7\pi/6)=-1$, giving $y=-2-(\sqrt3/2)x$. At $Q$, $\cos(11\pi/6)=\sqrt3/2$ and $2\sin(11\pi/6)=-1$, giving $y=(\sqrt3/2)x-2$.