The green line below is tangent drawn at point $P$. This construction uses the fact that the angle $PFT$ is $90$ degrees. But it doesn't give any explanation of why it must be $90$. Is there any simple way to see this with or without calculus?

Intuitively, when $P$ are right above $F$, it is clear that the angle is 90 because $PF$ is vertical and $FT$ is horizontal. As $P$ moves to the right, it seems $T$ also moves upkeep the angle at focus $90$. I'm not that sure how to approach proving things like these... Help appreciated.

enter image description here

  • $\begingroup$ Video on above construction is here . (Watching this video is not needed to answer my question. I think I've included all the necessary details in the question itself) $\endgroup$
    – AgentS
    Jun 28, 2018 at 5:20
  • 1
    $\begingroup$ I would look for similar or even congruent triangles. If $Q$ is the foot of the altitude from the directrix to $P$, then $PQ=PF$ and the tangent line bisects $\angle{FPQ}$. $\endgroup$
    – amd
    Jun 28, 2018 at 5:35
  • $\begingroup$ Ok.. In triangles $PFT$ and $PQT$ we have $\angle PQT=90$, $PQ = PF$ and $PT=PT$. How do we know the tangent bisects $\angle FPQ$ ? @amd $\endgroup$
    – AgentS
    Jun 28, 2018 at 5:49
  • $\begingroup$ The reflective property. $\endgroup$
    – amd
    Jun 28, 2018 at 6:57

1 Answer 1


Here's an image that can help you visualize the reflective property (incoming horizontal rays are reflected toward the focus) and why the tangent bisects $\angle FPQ$.

Tangent in P bisects angle FPQ

  • $\begingroup$ Ah, if we assume that the parabola reflects the incoming horizontal rays through the focus, then by angle of incidence = angle of reflection, the tangent indeed bisects $\angle FPQ$. Thank you :) the graph really helped :) I'll google a bit why the horizontal rays have to go through the focus. Thanks again :) $\endgroup$
    – AgentS
    Jun 28, 2018 at 10:22

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