Number of inscribed triangles in a rectangular hyperbola touching a parabola

Question

How many triangles can be incribed in the rectangular hyperbola $xy= c^2$ whose sides all touch the parabola $y^2 =4ax$.

How can we start the question . Please help.

Answer 1

My interpretation is that we should choose three different points $$\left(u,{c^2\over u}\right),\quad\left(v,{c^2\over v}\right),\quad\left(w,{c^2\over w}\right)$$ on the hyperbola $xy=c^2$ such that the three lines determined by these points are all tangent to the parabola $y^2=4ax$, $a>0$.

The line through the first two points has equation $$v(uy-c^2)=(u-x)c^2\ .$$ Intersecting it with the parabola leads to the quadratic equation $$v(uy-c^2)=\left(u-{y^2\over 4a}\right)c^2\ ,$$ or $${c^2\over 4a} y^2+uv\>y-c^2(u+v)=0\ .$$ For tangency the discriminant of this equation should be $0$, which leads to $$u^2v^2=-{c^4\over a}(u+v)\ .$$ In all we obtain three equations of this kind. In order to get rid of the parameters we write $$u:=\lambda \xi,\quad v:=\lambda\eta,\quad w:=\lambda\zeta$$ with $\lambda:=\bigl(c^4/a\bigr)^{1/3}$. We then have the three equations $$\xi^2\eta^2=-(\xi+\eta),\quad \eta^2\zeta^2=-(\eta+\zeta),\quad \zeta^2\xi^2=-(\zeta+\xi)\ .\tag{1}$$ Consider $\zeta\in\dot{\mathbb R}$ as given for the moment. Then the second and the third of these equations state that $\xi$ and $\eta$ are the solutions of one and the same quadratic equation. Without loss of generality we therefore have $$\xi={-1+\sqrt{1-4\zeta^3}\over 2\zeta^2},\quad \eta={-1-\sqrt{1-4\zeta^3}\over 2\zeta^2}\ .\tag{2}$$ With $(2)$ the second and third equation $(1)$ are fulfilled. We now have to make sure that the first equation is fulfilled as well, and this should give us conditions for $\zeta$. One obtains $$\xi+\eta=-{1\over\zeta^2},\quad\xi^2={2-4\zeta^3-2\sqrt{1-4\zeta^3}\over 4\zeta^4},\quad \eta^2={2-4\zeta^3+2\sqrt{1-4\zeta^3}\over 4\zeta^4}$$ so that we now have to ensure that $$(2-4\zeta^3)^2-4(1-4\zeta^3)=16\zeta^6\ .$$ But now a miracle happens: This is fulfilled identically in $\zeta$. Therefore we may choose any $\zeta\in\dot{\mathbb R}$ such that $4\zeta^3<1$ and obtain a configuration of the desired kind. So the answer to your question is $\infty\>$.

Answer 2

One way to start such a problem is "deflating the problem" or "elimination of technical terms" or "going back to definitions." (The phrases and the concept come from the section "Definitions" in George Polya's classic book How to Solve It.)

You have two words whose definitions are not clear: "inscribed" and "touch." Find ways to define these terms and reword your problem accordingly. Your problem is especially difficult since there are multiple possible ways to define these terms and each definition results in a different problem.

"Touch" probably means that each side of the triangle is to be tangent to the parabola. Clarify to yourself what that means in a way that can help solve the problem. For example, you could choose three points on the parabola, extend the lines tangent to the parabola through those points, and consider the triangle formed by those three lines.

That should get you started. If you need more help, tell us appropriate definitions of "inscribed" and "touch" and we may be able to help you a little further. Without those definitions, your question is too vague and will probably be closed.