Show that $PF.PG=b^2$ in a hyperbola

Question

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.

Answer 1

This isn't a "pure geometric solution" since I've used some results from analytic geometry, but it does save you all the hassle of finding co-ordinates and calculating slopes and distances.

All the $\color{green}{green}$ angles in the figure are equal to $\phi$ , where $\tan\phi={b\over a}\csc\theta$.

If you use a trigonometric identity you'll get $$\cos^2\phi={a^2 \over a^2+b^2\csc^2\theta}$$ This will be of use later on. Here $\theta$ is the eccentric angle of point $P$.

PROOF#1

Since $CDPF$ is a rectangle, $CD=PF$. Also since $\Delta CDO $~$\Delta PNG$, you have: $${CD \over PN}={OC \over PG}$$ $$PG\cdot CD=OC\cdot PN$$ $$PG\cdot PF=OC\cdot PN$$ Now as we know that the point $P$ is at ($a\sec\theta$,$b\tan\theta$)$\implies$ $PN=b\tan\theta$

$OC$ is simply the y-intercept of the tangent at $P$ $\implies$$OC=b\cot\theta$

This gives $\color{red}{PF\cdot PG}=b\tan\theta*b\cot\theta=\color{red}{b^2}$.

PROOF#2

Applying simple trigonometry, $PF=CD=OC\cos\phi$ and $PG'=OG'\cos\phi$. Substituting these values in $PF\cdot PG'$: $$PF\cdot PG'=OC\cdot OG'\cdot \cos^2\phi$$

Now $OC$ is simply the y-intercept of the tangent and $CG'$ is simply the y-intercept of the normal, so we write:

$OC=b\cot\theta$ and $CG'={a^2+b^2 \over b}\tan\theta$. Also note that $OG'=OC+CG'$.

Substituting these values in $OC\cdot OG'$,

$$OC\cdot OG'=b\cot\theta*\left( b\cot\theta+ {a^2+b^2 \over b}\tan\theta \right)=a^2+b^2+b^2\cot^2\theta=a^2+b^2\csc^2\theta$$ What do you get by multiplying it with $\cos^2\phi$?

Answer 2

$\rm PF.PG$ = 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, $${\rm PF.PG}=|a^2\sec^2\theta+b\tan\theta(b\tan\theta-(1/b)(a^2+b^2)\tan\theta)|=a^2$$