# 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.

• i wish to answer this but it's too long – RE60K Feb 27 '15 at 14:46
• @Abhishek please use \ for all trig functions like \cos in latex. – Narasimham Feb 27 '15 at 14:57
• @Narasimham, \cosec doesn't work.... – Abhishek Bakshi Feb 27 '15 at 17:36
• @ADG, I had the experience, that's why I am asking for a easier approach.. – Abhishek Bakshi Feb 27 '15 at 17:37
• use \csc for cosec and \left( and \right) for bigger brackets – RE60K Feb 27 '15 at 17:40

## 2 Answers

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 ($a\sec\theta$,$b\tan\theta$)$\implies$ $PN=b\tan\theta$