<s>[INST] Show that $PF.PG=b^2$ in a hyperbola If the normal at P to the hyperbola $\frac {x^2}{a^2}-\frac {y^2}{b^2}=1$ meets the transverse axis in G and the conjugate axis in G' and CF be the perpendicular to the normal from the center C then show that $$PF.PG=b^2\space and \space PF.PG'=a^2.$$ We know that the equation of the normal at parametric point $P\equiv(a\sec\theta,b\tan\theta)$ is given by
$$a\,x\cos\theta+b\,y\cot\theta=a^2+b^2$$ 
Equation of the transverse axis is $y=0$ and that of the conjugate axis is $x=0$. 
Hence 
$$G\equiv(\frac {a^2+b^2}{a}\sec\theta,0), \, G'\equiv(0,\frac {a^2+b^2}{b}\tan\theta)$$
C is the origin and CF is perpendicular to the normal. Hence 
$$ CF\equiv b\,x\sec\theta-a\,y\tan\theta=0\implies y=\frac {b\,x\sec\theta}{a \tan\theta}$$ 
Substituting this in the equation of the normal we get 
$$a\,x\cos\theta+b \,\cot\theta\,\frac {b\,x\sec\theta}{a\,\tan\theta}=a^2+b^2\ $$ 
implies
$$ (a^2 \cos\theta \sin^2\theta+ b^2\cos\theta)x=(a^2+b^2)\sin^2\theta$$
After finding the coordinates of F, the calculation becomes very complicated. So is there an easier way to approach the problem? I would love to see a pure geometric solution to this problem.
 [/INST] $\rm PF.PG$ = Power of P wrt the circle with CG as diameter, the equation of whose is:
$$x(x-(1/a)(a^2+b^2)\sec\theta)+y^2=0$$
So, $${\rm PF.PG}=|a\sec\theta(a\sec\theta-(1/a)(a^2+b^2)\sec\theta)+b^2\tan^2\theta|=b^2$$
Similiarly:
$\rm PF.PG'$ = Power of P wrt the circle with CG' as diameter, the equation of whose is:
$$x^2+y(y-(1/b)(a^2+b^2)\tan\theta)=0$$
So, $${\rm PF.PG}=|a^2\sec^2\theta+b\tan\theta(b\tan\theta-(1/b)(a^2+b^2)\tan\theta)|=a^2$$
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