Poincaré upper half plane model

The Poincaré upper half plane model for $\mathbb{H}^2$ is the upper half plane $\{(x,y) \in \mathbb{R}^2 : y>0 \}$ in which a point is similar to the Euclidean point and a line must be one of the following:

• a vertical ray (excluding its endpoint) extending from the boundary;
• an semicircle (excluding its endpoints) whose center lies on the boundary.

The Poincaré upper half plane model has the drawback that lines in the model do not necessarily resemble Euclidean lines; however, it has the advantage that it is angle preserving. That is, the Euclidean measure of an angle within the model is the angle measure in hyperbolic geometry. This model has the added bonus that analytic geometry is a useful tool for performing constructions. For example, consider the following:

Solution: The common perpendicular cannot be a vertical ray, so it must be a semicircle. Also, if a semicircle is to be perpendicular to $x=-4$ , it must have a center at $(-4,0)$ . Thus, the common perpendicular is of the form $y=\sqrt{r^2-(x+4)^2}$ for some $r>0$ .

Since $y=\sqrt{r^2-(x+4)^2}$ must also be perpendicular to $y=\sqrt{6+x-x^2}$ , their tangent lines at their point of intersection must be perpendicular. Let $(x_0,y_0)$ denote this point of intersection. Thus, the line tangent to $y=\sqrt{6+x-x^2}$ at $(x_0,y_0)$ must pass through $(-4,0)$ . Let $m$ denote the slope of this line. Then $\displaystyle m=\frac{y_0}{x_0+4}=\frac{\sqrt{6+x_0-(x_0)^2}}{x_0+4}$ .

For $y=\sqrt{6+x-x^2}$ , $\displaystyle \frac{dy}{dx}=\frac{1-2x}{2\sqrt{6+x-x^2}}$ . Thus, $\displaystyle \frac{\sqrt{6+x_0-(x_0)^2}}{x_0+4}=\frac{1-2x_0}{2\sqrt{6+x_0-(x_0)^2}}$ . Solving for $x_0$ yields:

Now $y_0$ can be found:

Finally, $r^2$ can be found:

Hence, the common perpendicular is $\displaystyle y=\sqrt{\frac{1117}{81}-(x+4)^2}$ .