## Locus problems related to the chord of a conic

Pavel Pech, Roman Hašek
University of South Bohemia

Example 1: Let $UV$ be a chord of an ellipse centered at $O$ such that ${OU}\perp{OV}$. Determine the locus of the foot $P$ of the perpendicular from $O$ to the chord $UV$ when $U$ moves along the ellipse.

The appropriate locus is a circle.

Experimenting with the position of point O it turns out that it may not be at the center of a conic.

This enables us to apply this construction to all conics, e.g. to a parabolla, as given below.

It also works for singular conics.

If the lines are mutually orthogonal, we get a line instead of a circle.

Proof:  The case of two orthogonal lines can be proved straightforwardly. Applying the Simson-Wallace theorem on the triangle QUV and O on its circumcircle, the points P, S, R are collinear (the Simson line).

Example 2: (Frégier point) Given a conic κ and a point O on κ, then the hypotenuses of right-angled triangles inscribed to κ and having common right-angle vertex O intersect at one point F – the Frégier point to O with respect to κ.

Proof:  When U moves along κ, the point P forms a circle c. Since the angle OPU is right, then also the angle OPF is right. Then according to the theorem of Thales the point F must be fixed. It is the opposite point to the point O on the circle c.

Example 3: Given a line k, a point O on it, let the line l be perpendicular to k at O with a point M on it, and let A be an arbitrary point in the plane. Determine the locus of the intersection P of the circle c centered at M with radius MO with the line AB, where B ∈ l is the opposite point to O in the circle c, when M moves along the line l.

If M moves along the line l then the locus of P is a circle. Try it using the dynamic figure below.

Proof:  According to the Thales' theorem, the angle BPO is right. This implies that the angle APO is right as well. Using the Thales' theorem once again, we come to the conclusion that the point P lies on a circle with a fixed diameter |OA|.