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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.)

enter image description here

It seems I've proved that the point can't bisect the chord, which sounds contradictory. I guess something is wrong in my proof.

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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 $y_C=(y_A+y_B)/2$.

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  • $\begingroup$ But what's wrong in my solution? Why am I getting a different answer? Sorry, but I'm currently learning high school mathematics and we haven't been told anything about hyperbolic functions. :( $\endgroup$ – Shivansh J Mar 20 '18 at 10:46
  • $\begingroup$ Your solution doesn't work because $\alpha$ and $\beta$ are not independent among them: you didn't take into account the equation for $\sec\theta$. It's simpler, in my opinion, to find the intersections between the tangent line and the other hyperbola. $\endgroup$ – Aretino Mar 20 '18 at 11:31
  • $\begingroup$ See also my edit. $\endgroup$ – Aretino Mar 20 '18 at 11:54
  • $\begingroup$ Thanks your solution is absolutely correct, but I'm still not able to understand what went wrong with my method, can you please elaborate a bit more. Well, you said 'you didn't take into account the equation for sec (theta).' What does it mean? When I assumed the coordinates for conjugate hyperbola involving variable theta, (parametric form) it automatically implies that the point will have to lie on that curve, moreover the tangent to the conjugate hyperbola is the chord of standard hyperbola, so equation of tangent and chord will have to be same. $\endgroup$ – Shivansh J Mar 20 '18 at 18:10
  • $\begingroup$ What do you mean by saying "alpha and beta are not independent among them"? $\endgroup$ – Shivansh J Mar 20 '18 at 18:14
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This is proven by Apollonius of Perga (about a half century after Euclid) in Conics II, 19. "If some straight line is drawn touching some one of the conjugate opposite sections at random, it will meet the adjacent sections and will be bisected at the point of contact." This was of course before coordinate geometry. Apollonius' proof uses no trigonometry, but rests on earlier propositions in Conics I & II, which in turn rest on Euclid's Elements.

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