What would be a good approach to tackle this problem. In a previous assignment I managed to show Pq=Pr. How do I show that this tangent intersects the conjugate hyperbola. Should I start by calculating the arbitrary tangent line and seeing how it intersects the conjugate and then show that Qq=Rr?
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The solution:
Well I like to work with: $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$ Let point P be $(a\sec\theta,b\tan\theta)$ and the tangent and pair of asymptotes is: $$\frac xa\sec\theta-\frac yb\tan\theta=1\\ \frac{x^2}{a^2}=\frac{y^2}{b^2}\implies y=\pm\frac bax$$ Now I am finding intersection: $$\frac xa\sec\theta\pm\frac xa\tan\theta=1\implies x=a(\sec\theta\pm\tan\theta)\tag{q and r}$$ These two point correspond to $q$ and $r$: $$q\equiv(a(\sec\theta+\tan\theta),b(\sec\theta+\tan\theta))\\ r\equiv(a(\sec\theta-\tan\theta),-b(\sec\theta-\tan\theta))$$ Now conjugate hyperbola is: $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$$ Again finding intersection: $$x=a(sec\theta\pm\sqrt2\tan \theta)\tag{Q and R}$$ So Q and R are: $$Q\equiv\left(a\left(\sec \theta+\sqrt2\tan \theta\right),b\left(\tan \theta+\sqrt{2}\sec \theta\right)\right)\\ R\equiv\left(a\left(\sec \theta-\sqrt{2}\tan \theta\right),b\left(\tan \theta-\sqrt{2}\sec \theta\right)\right)$$ Clearly $PQ=PR,Qq=Rr,qR=Qr$ using distance formula and the fact that $\alpha^2=(-\alpha)^2$.