Here are seven results about parabolas, of the sort often proved by laborious algebra for A level exams, proved from a single diagram by geometry. Results 2 and 5 are also proved, but are important as steps to prove the other results rather than for themselves.

For once, by the way, it’s not just Edexcel to blame for the turn from elegant geometry to brute-force algebra. By chance I found demands for algebraic proofs of a couple of these results in a Cambridge Tripos paper of 1840.

For simplicity, the diagram omits the actual parabola itself. P and Q are points on a parabola. F is the focus. DE is the directrix. PT and QT are the tangents at P and Q.

By the focus-directrix definition of a parabola, FP=DP and FQ=EP.

1. TP bisects ∠ FPD (and similarly TQ bisects ∠ FQD)

Proof: The line bisecting the angle is the line with points equally distant from F and D. By definition of a parabola, the tangent at P, i.e. the line most closely approximating the parabola, will also have its points equally distant from F and D. Therefore the tangent and the line bisecting the angle are identical. (Click here for a more formal argument).

2. PDTF is a kite (and so is QETF)

Proof: since TP bisects ∠ FPD, and FP=DP, TP is the perpendicular bisector of FD.

3. Every ray travelling parallel to the axis of a parabolic mirror is reflected by that mirror to the focus F

Proof: The broken line PN is the normal at P. What we need to prove is that ∠ XPN = ∠ FPN.

∠ FPN + ∠ FPT = π/2 (because normal ⊥ tangent)

∠ XPN + ∠ FPN + 2 × ∠ FPT = π (straight line)

Therefore ∠ XPN = ∠ FPN.

4. TF bisects ∠ PFQ

Proof: Since DT = TF (kite) and ET = TF (kite), △ DTE is isoceles.

∠ TDE = ∠ TED = γ, say

∠ TFP = ∠ TDP = π/2 − γ

Similarly ∠ TFQ = π/2 − γ, therefore ∠ TFP = ∠ TFQ

5. The two kites are similar

Proof: ∠ DTE = π − 2 γ

Adding up all the angles round the point T,

∠ DTE + 2 × ∠ DTP + 2 × ∠ ETQ = 2π

Halving, π/2 − γ + ∠ DTP + ∠ ETQ = π

But from triangle TDP, π/2 − γ + ∠ DTP + ∠ DPT = π

Therefore ∠ DPT = ∠ ETQ

By a similar argument ∠ EQT = ∠ DTP

Therefore the angles of the two kites are identical, therefore the kites are similar.

6. PF.QF = TF^{2}

Proof: By similarity PF/TF = TF/QF (the ratio of the longer side to the shorter side is the same in both kites).

7. T is halfway between DP and EQ (or y_{T} = ½ (y_{P} + y_{Q}))

Proof: Triangle DTE is isosceles.

8. The foot of the line from the focus to a tangent lies on the tangent at the vertex (or: R and S are on the tangent at the vertex)

Proof: R is halfway along the line FD because PDTF is a kite, and so it is also halfway between F and DE horizontally, i.e. on the line parallel to the directrix going through the vertex, which is the tangent at the vertex. This also shows y_{R} = ½ (y_{F} + y_{P}).

9. T lies on the directrix if and only if PT ⊥ QT

Proof: That happens if and only if γ = 0 and ∠ DTE = π, in which case adding all the angles round T tells us:

π + 2 × ∠ DTP + 2 × ∠ ETQ = 2π

i.e. ∠ DTP + ∠ ETQ = π/2

∠ FTP + ∠ FTQ = π/2; but ∠ PTQ = ∠ FTP + ∠ FTQ.

If P and Q are both on the same side of the axis of the parabola, the diagram is different in that the kites are arrow-shape rather than ordinary kite-shape, but the argument is the same.