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).
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.
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?