The image below shows the 'AOL' building in Dublin - Leaving Cert DCG - Question A-2 - 2019

To find the three points outside the latus rectum, we can consider points that lie beyond the hyperbola, following the trajectory of the curve above the directrix. Possible coordinates are (3, y₁), (4, y₂), and (5, y₃), where y₁, y₂, and y₃ are values calculated from the hyperbola's equation.

For the two points located inside the latus rectum, we will choose points that lie on the inner part of the hyperbola, specifically points near the focus F, such as (1, y₄) and (2, y₅), calculated using the curve's equation.

To draw the top portion of the hyperbola, we can plot the vertex A and the identified points, using smooth curves to connect them illustrating the hyperbolic shape, while ensuring the branches properly extend outward at the foci.

At point P, construct a line that is perpendicular to the eccentricity line by first finding the slope of the line connecting the focus F to P, and then taking the negative reciprocal of that slope to draw the perpendicular line.

To find the center, draw two segments: one from the focus F to the point where the eccentricity line meets the directrix, and another from F to point P. The angle bisector of these two lines will indicate the center of the hyperbola.

The required circle can be drawn with its center located at the point of intersection from the bisector drawn earlier. The radius should be determined such that the circle is tangent to both the hyperbola at point P and the directrix DD₁.