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Let S be the focus of the hyperbola $xy=1$. Let a tangent to the hyperbola at point P cuts the latus rectum (through S) produced, at point Q and the directrix (corresponding to S) at point T. Also let M be the foot of perpendicular drawn from the point P to the same directrix. If angle PTS=$\theta_1$ and angle PMS=$\theta_2$, find $\frac{\theta_1}{\theta_2}$ and $\frac{SQ}{ST}$

My Attempt:

I made the diagram and guess that P,M,T,S lies on a circle. Not sure though.

Tangent at P is $\frac{x}{x_1}+\frac{y}{y_1}=2$, where $(x_1,y_1)$ are the coordinates of P.

Taking x-axis as the directrix, PM=$y_1$, MT=$√2-x_1$

Taking S as $(√2,√2)$

Not able to proceed ahead.

aarbee
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    HINT 1. Points $PMTS$ do lie on a circle, because of this:https://math.stackexchange.com/questions/2360164/a-right-angle-at-the-focus-of-a-hyperbola – Intelligenti pauca Mar 26 '23 at 16:04
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    HINT 2. Triangles $MPS$ and $QST$ are similar. – Intelligenti pauca Mar 26 '23 at 16:05
  • @Intelligentipauca, thanks for the hints. I am able to work out Hint1. For Hint 2, I can see one pair of equal angles. Not able to prove triangles to be similar. – aarbee Mar 27 '23 at 05:46
  • HINT 3. $PS\perp ST$ and $PM\perp SQ$, hence $\angle MPS =\dots$. – Intelligenti pauca Mar 27 '23 at 13:24
  • @Intelligentipauca sorry not able to take the hint. Referring to the diagram below (Narasimham's answer), I can see that PS perpendicular to ST and PM perpendicular to SQ. Not able to see the relation with angle MPS here. – aarbee Mar 27 '23 at 17:23
  • Rotate triangle $QST$ by $90°$, so that $Q$ is mapped to $Q'=M$: line $Q'S'$ is then the same as line $MP$ and line $S'T'$ parallel to $PS$. We have then $\angle MPS=\angle Q'S'T'=\angle QST$. – Intelligenti pauca Mar 27 '23 at 18:19
  • @Intelligentipauca sorry to press further but I am still not able to see it :-(. On roation, yes, the way diagram appears, 'looks' like angle MPS = angle Q'S'T' but I am not able to see why this is happening. e.g. I can see that MP is parallel to Q'S' because earlier they were perpendicular and now we have rotated them by 90 degree. But I don't see why PS will be parallel to S'T', though it does appear parallel in the diagram. – aarbee Mar 28 '23 at 01:49

2 Answers2

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We have $\angle PMT=90°$ by construction and $\angle PST=90°$ by a well-known property of any conic section. Hence quadrilateral $PSTM$ is cyclic and $\angle PMS=\angle PTS$.

Line $QS$ is perpendicular by construction to $PM$, while $PS\perp ST$ as seen above. Hence $\angle QST=\angle SPM$ because the sides of these angles are pairwise perpendicular. It follows that triangles $QST$ and $SPM$ are similar and $SQ/ST=PS/PM=e=\sqrt2$.

enter image description here

Intelligenti pauca
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  • Thanks for the answer. If the sides of the angles are pairwise perpendicular then angles will be equal? I am not aware of this. I searched this on google, not getting any specific result. Maybe I need to search with different keywords? – aarbee Mar 28 '23 at 01:53
  • @aarbee Let's try another way. Produce $SQ$ to meet line $PM$ at $Q'$. Circle $PSQ'$ has diameter $SP$, because $\angle SQ'P=90°$. But $ST\perp SP$, hence $ST$ is tangent to the circle. It follows that $\angle Q'PS=\angle Q'ST$ as both inscribed angles insist on the same arc. – Intelligenti pauca Mar 28 '23 at 11:03
  • I think I understood this time. Are we talking about Alternate Segment Theorem? – aarbee Mar 28 '23 at 20:23
  • @aarbee Yes, exactly. – Intelligenti pauca Mar 28 '23 at 20:45
  • Thank you for your time. – aarbee Mar 28 '23 at 21:21
  • One doubt. If I make the diagram with point P lower than point Q, then I am not able to apply the alternate segment theorem. – aarbee Mar 29 '23 at 07:15
  • @aarbee Everything works even in that case, with $Q'$ intersection between lines $SQ$ and $PM$. Just some little changes, which should be evident from the diagram. – Intelligenti pauca Mar 29 '23 at 11:46
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Rotating the hyperbola through $-45^{\circ}$ the classic (rectangular) hyperbola with symmetry along axes is obtained and believe easier to work with further on with a rough sketch.

Eccentricity $e$, QS on latus rectum, DTM directrix, OD= $a/e$, $OS=a e,~ $ from y-axis.

$$~ a=1; ~ x^2- y^2= 2 a^2 = a^2 e^2;e = \sqrt{2} ;\tag1 $$

A very useful property of conics that may be applied here. Newton's conic, x-axis horizontal. Standard geometric symbols.

$$ r(1- e \cos \theta)=p; ~~ (r- e \cdot x) =p;$$ Differentiate w.r. arc of conic and use differential relations.

$$ \cos \psi -e \cos \phi =0 $$

$$\boxed{ \frac{\cos \psi}{\cos \phi}= e }\tag 2 $$

We have from cyclic quadrilateral $PMTS$ two angles sum at M

$$ \theta + \psi = \pi/2 \tag 3 $$ and finally

$$ \theta_1=\theta_2=\theta, ~~ \frac{SQ}{ST}=\frac{\sin \theta}{\sin{(\pi/2 -\phi)}} =\frac{\sin \theta }{\cos \phi} $$

$$ = \frac{\cos \psi}{\cos \phi} = e= \sqrt {2}\tag 4 $$

that completes the proof.

It is verified ( same angle subtended on a point opposite a segment) there are three independent angles in any cyclic quadrilateral, and the fourth dependent angle $\pi- ( \theta+ \psi+ \phi ) $

enter image description here

Narasimham
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  • Hey, thanks for the hint. I can see that $\theta_1=\theta_2$. Not able to find $\frac{SQ}{ST}$ – aarbee Mar 27 '23 at 05:47
  • Ok. my sketch is hand drawn Paint but suggest a better sketch would on Geogebra, desmos etc. graphing softwares using rotated axes. – Narasimham Mar 27 '23 at 11:30
  • Hey, the sketch is big help. I was just looking for a hint to find the ratio of SQ and ST. – aarbee Mar 27 '23 at 12:11
  • You know the equation of tangent and where it cuts directrix & latus rectum. Added another rough sketch but here also a graph mistake. The vertical lines should be ( $ x=1, x=2 $) – Narasimham Mar 27 '23 at 12:24
  • Yes, that is true. I can find the ratio as the coordinates are known. Thanks. The hint above by Intelligenti Pauca seems to suggest we need not use distance formula. If triangles are similar then the ratio is just the eccentricity. But I am not able to prove the similarity of triangles. I can see only one pair of equal angles i.e. $\theta_1=\theta_2$ – aarbee Mar 27 '23 at 13:10
  • Thanks for the edit. I am aware that the ratio of focal distance to the distance from directrix is eccentricity. But I am not aware of the ratio involving cosines. Do you mind giving any link please? – aarbee Mar 27 '23 at 17:28
  • I have included derivation which leads to the final proof. – Narasimham Mar 27 '23 at 19:11