I have a question that says the following:

A line thought the origin meets the circle $x^2+y^2=a^2$ at P and the hyperbola $x^2-y^2=a^2$ at Q. What is the locus of the point of intersection of the tangent at P to the circle with the tangent at Q to the hyperbola?

So, the few things I noticed right away were, clearly the given circle is the auxiliary circle to the hyperbola. So, what I did was, I tried to take the points P and Q in parametric form and write equation of tangents to the circle and hyperbola using that, but I hit a snag:

Can I use the same parametric angle for both P and Q, i.e, "$\theta$" ??

Or do I need to use two different parametric angles for P and Q, in which case, I think I would be better off assuming the equation of the line through the origin as $y = mx$ and then finding P and Q by solving this equation of the line and the equations of the curves simultaneously and then find the tangents at those points? Is this approach correct?

  • $\begingroup$ "Can I use the same parametric angle for both P and Q, i.e, "$\theta$" ??" If you take $P=(a\cos\theta,a\sin\theta)$ and $Q=(a\sec\theta,a\tan\theta)$, then (almost-always) $\frac{P_y}{P_x}\neq \frac{Q_y}{Q_x}$, so these points are (almost-always) not on a common line through the origin which makes them not the points you need. Thus, the answer to that question is "No". ... Your alternatives are sound, but it might be easiest (I haven't thought it all the way through) to take $P$ as above and $Q=k(\theta)\;P$, with function $k(\theta)$ determined by having $Q$ satisfy the hyperbola equation. $\endgroup$
    – Blue
    Sep 15 '20 at 5:06
  • $\begingroup$ @Blue Which points do you mean, are almost always not on the common line through origin? $\endgroup$
    – Techie5879
    Sep 15 '20 at 5:08
  • $\begingroup$ I mean exactly those points $P$ and $Q$ that I parameterized using the same parametric angle $\theta$. That is, if $P$ is the point on the circle determined using (central) angle $\theta$, then the point on the hyperbola determined using (auxiliary) angle $\theta$ is (almost-always) not the "$Q$" in the problem. (They match when $\theta=0$, so I can't say "always". :) $\endgroup$
    – Blue
    Sep 15 '20 at 5:19
  • 1
    $\begingroup$ @Blue Yes thanks for clarifying, that is exactly what I was thinking. Thanks again!! $\endgroup$
    – Techie5879
    Sep 15 '20 at 5:20

Let the line throught origin be $y=mx$ its intersection with the circle is the point $P_1(a/\sqrt{1+m^2},ma/\sqrt{1+m^2})$. Its point of intersection with the hperbola is $P_2(a/\sqrt{1-m^2}, ma/\sqrt{1-m^2}).$ Thangen $T_1$ at $P_1$ is $$x+my=a\sqrt{1+m^2} ~~~(1)$$ Similarly the tangent $T_2$ at $P_2$ is $$x-my=a\sqrt{1-m^2}~~~(2)$$ adding them we get $$2x=a[\sqrt{1+m^2}+\sqrt{1-m^2}]~~~~~(3)$$ subtracting them we get $$2my=a[\sqrt{1+m^2}- \sqrt{1-m^2}]~~~~(4)$$ Multiplying (3) and (4), we get $$4mxy=2a^2 m^2 \implies m=2xy/a^2.$$ Putting this in (3), we get the eliminant as $$2ax=\sqrt{a^4+4x^2y^2}+\sqrt{a^4-4x^2y^2} ~~~(5)$$ This gives the required locus.


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