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?
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.