While studying coordinate geometry I came across a question -
Prove that chords of a hyperbola, which touch the conjugate hyperbola, are bisected at the point of contact.
Since everyone hyperbola can be transformed into a standard hyperbola, whose transverse and conjugate axis are along the coordinate axis by suitable shifting of origin, I can prove this question for a standard hyperbola.
Unfortunately, in spite of many tries, I'm not able to prove this statement. Any help will be greatly appreciated.
Here's my try, it includes the figure (in case if statement is not clear, also note that the figure is not to scale.)
It seems I've proved that the point can't bisect the chord, which sounds contradictory. I guess something is wrong in my proof.
The upper branch of the conjugate hyperbola can be described by the following parametric equations: $$ x=a\sinh t \\ y=b\cosh t \\ $$ Consider now a generic point $P=(a\sinh\bar t,b\cosh\bar t)$ on that hyperbola: the equation of the line tangent at $P$ (with parameter $s$) is then $$ x=(a\cosh\bar t)s+a\sinh\bar t \\ y=(b\sinh\bar t)s+b\cosh\bar t \\ $$ To find the intersections of this line with the hyperbola, we must insert $x$ and $y$ given above into the equation $x^2/a^2-y^2/b^2=1$ and then solve for $s$: this gives the simple result $s^2-1=1$, that is: $s=\pm\sqrt2$.
We have then two intersection points, whose coordinates are obtained by adding and subtracting the same quantities to the coordinates of $P$, which is their midpoint.
EDIT.
With your parameterization, the equation of the tangent line gives $$ {y\over b}=\cos\theta+{x\over a}\sin\theta $$ and substituting that into the equation $(x/a)^2-(y/b)^2=1$ of the other hyperbola one readily gets the coordinates of the intersection points $A$ and $B$: $$ x_{A/B}=a\tan\theta\pm\sqrt2 a\sec\theta, \quad y_{A/B}=b\sec\theta\pm\sqrt2 b\tan\theta. $$ As you can check, $x_C=(x_A+x_B)/2$ and $y_C=(y_A+y_B)/2$.