# Two Tangents to Parabola

Assume there are two points A and B on a parabola, with tangents AS and BS meeting in S. A' and B' are the feet of perpendiculars from A and B to the directrix of the parabola. Then

S is the circumcenter of ΔA'B'F,

∠FAS = ∠FA'B' = ∠FSB,

∠FBS = ∠FB'A' = ∠FSA,

Triangles BFS, SFA, and B'FA' are similar,

F lies on AB iff S lies on A'B' and, in this case, ∠ASB = 90

^{o}.

### Proof

From the reflective properties of parabola, SA is the perpendicular bisector of FA'. Therefore,

SA' = SF. Similarly,SB' = SF. An inscribed angle FA'B' and a central angle FSB' subtend the same chord. Hence,

∠FSB = ∠FSB'/2 = ∠FA'B'.

On the other hand, angles FA'B' and SAA' have pairwise perpendicular legs and are thus equal. Since, ∠SAA' = ∠FAS,

∠FA'B' = ∠FAS.

Similar to 2.

Follows from 2-3. Triangles BFS, SFA, and B'FA' have the same orientation and are directly similar. Triangles SA'A and BB'S are also similar the above three, but have a different orientation.

To shorten the expressions, let's introduce

α = ∠FAS andβ = ∠FBS. If S lies on the directrix,2(α + β) = 180°. Then∠AFS = (90° - α) + (90° - β) = 90°, and similarly∠BFS = 90°. So AFB is a straight line. The argument is obviously reversible.

The latter property is often referred to as **orthoptic**:

Two tangents to a parabola are perpendicular iff their point of intersection lies on the directrix.

### References

- H. Dörrie,
*100 Great Problems Of Elementary Mathematics*, Dover Publications, NY, 1965

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