4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ i wish to answer this but it's too long $\endgroup$ – RE60K Feb 27 '15 at 14:46
  • $\begingroup$ @Abhishek please use \ for all trig functions like \cos in latex. $\endgroup$ – Narasimham Feb 27 '15 at 14:57
  • $\begingroup$ @Narasimham, \cosec doesn't work.... $\endgroup$ – Abhishek Bakshi Feb 27 '15 at 17:36
  • $\begingroup$ @ADG, I had the experience, that's why I am asking for a easier approach.. $\endgroup$ – Abhishek Bakshi Feb 27 '15 at 17:37
  • $\begingroup$ use \csc for cosec and \left( and \right) for bigger brackets $\endgroup$ – RE60K Feb 27 '15 at 17:40
1
$\begingroup$

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.

enter image description here

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 $\cos^2\phi$?

$\endgroup$
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.