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I'd like to prove the reflection property for the hyperbola. That is, that S'PS is bisected by the tangent at P. Suppose the tangent intersects the x axis at T. The usual method would be to use the sine rule on triangles S'PT and SPT, then to use the fact thar TS/TS'=PS/PS' (can be shown quite quickly with an easy computation).

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I'm trying to find an alternative proof by constructing perpendiculars from S and S' to the tangent at R and R' respectively. However from here, I don't have much other than...

  • Triangles TSR and TR'S' are similar.
  • I need to somehow show that triangles SPR and S'PR' are similar.

Would love to receive any pointers to get me going in the right direction to an alternative proof.

Trogdor
  • 10,331

1 Answers1

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Well, you have $TSR$ and $TR'S'$ similar, hence $\dfrac{RS}{R'S'}=\dfrac{TS}{TS'}$

You have also, as you stated, $\dfrac{TS}{TS'}=\dfrac{PS}{PS'}$

Then you have $\dfrac{RS}{R'S'}=\dfrac{PS}{PS'}$

That is also $\dfrac{R'S'}{PS'}=\dfrac{RS}{PS}$.

Can you finish from here?

Martigan
  • 5,844