# Question on rectangular hyperbola and its focus and directrix.

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 $$(√2,√2)$$

• HINT 1. Points $PMTS$ do lie on a circle, because of this:math.stackexchange.com/questions/2360164/… Commented Mar 26, 2023 at 16:04
• HINT 2. Triangles $MPS$ and $QST$ are similar. Commented Mar 26, 2023 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. Commented Mar 27, 2023 at 5:46
• HINT 3. $PS\perp ST$ and $PM\perp SQ$, hence $\angle MPS =\dots$. Commented Mar 27, 2023 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. Commented Mar 27, 2023 at 17:23

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 $$SQ/ST=PS/PM=e=\sqrt2$$.

• 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? Commented Mar 28, 2023 at 1:53
• @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. Commented Mar 28, 2023 at 11:03
• I think I understood this time. Are we talking about Alternate Segment Theorem? Commented Mar 28, 2023 at 20:23
• @aarbee Yes, exactly. Commented Mar 28, 2023 at 20:45
• Thank you for your time. Commented Mar 28, 2023 at 21:21

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 $$\pi- ( \theta+ \psi+ \phi )$$

• Hey, thanks for the hint. I can see that $\theta_1=\theta_2$. Not able to find $\frac{SQ}{ST}$ Commented Mar 27, 2023 at 5:47
• Ok. my sketch is hand drawn Paint but suggest a better sketch would on Geogebra, desmos etc. graphing softwares using rotated axes. Commented Mar 27, 2023 at 11:30
• Hey, the sketch is big help. I was just looking for a hint to find the ratio of SQ and ST. Commented Mar 27, 2023 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 ( $x=1, x=2$) Commented Mar 27, 2023 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. $\theta_1=\theta_2$ Commented Mar 27, 2023 at 13:10