Show that an infinite number of triangles can be inscribed in either of the parabolas $y^2=4ax$ and $x^2=4by$ whose sides touch the other parabola.

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Show that an infinite number of triangles can be inscribed in either of the parabolas $y^2=4ax$ and $x^2=4by$ whose sides touch the other parabola.

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By applying an affinity $(x,y)\mapsto(\alpha x,\beta y)$ with some suitable constants $\alpha,\beta$ we can transform the two parabolas to $y^2=x$ and $x^2=y$, with keeping tangency.

Lemma: Suppose that $(u,u^2)$ and $(v,v^2)$ are two points on parabola $x^2=y$. Then the line connecting $(u,u^2)$ and $(v,v^2)$ is tangent to the parabola $x=y^2$ if and only if $uv(u+v)=\frac{-1}{4}$.

Proof: The slope of line connecting the two points is $\frac{v^2-u^2}{v-u}=u+v$, so its equation is $y-u^2=(u+v)(x-u)$. The line is tangent to $y^2=x$ if and only if the equation $y-u^2=(u+v)(y^2-u)$ is quadratic (i.e. $u+v\ne0$) and its discriminant is $0$: $$ (-1)^2 - 4\cdot (u+v) \cdot (-uv)=0. $$

For $u$ and $v$ we can find the third point: let $w=-u-v$; then we achieve $uw(u+w)=vw(v+w)=uv(u+v)=\frac{-1}4$, so all three sides of the triangle formed by the vertices $(u,u^2)$, $(v,v^2)$, $(w,w^2)$ are tangents to the parabola $x=y^2$.

Remark 1. By Poncelet's theorem, it is sufficient to find a single triangle with this property.

Remark 2. It also can be seen that $uvw=\frac14$, so $u,v,w$ are the roots of a cubic equation of te form $X^3+cX-\frac14=0$ with some $C$. This leads to the explicit form of the solutions: $$ u = \frac{\cos \varphi}{\sqrt[3]{\cos 3\varphi}}, \quad v = \frac{\cos (\varphi+\frac{2\pi}3)}{\sqrt[3]{\cos 3\varphi}}, \quad w = \frac{\cos (\varphi+\frac{4\pi}3)}{\sqrt[3]{\cos 3\varphi}} \quad (-\frac\pi3<\varphi<\frac\pi3) $$ and its permutations.