Question on rectangular hyperbola and its focus and directrix.

Question

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.

Answer 1

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

Answer 2

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 ) $