A family of lines normal to one and tangent to another curve

Question

For any real value of the parameter $\theta$, if the line $$y=\sin \theta (x-2) - 9 \sin \theta + \sin 3 \theta $$ is normal to one curve $f(x,y)=0$ and it touches another curve $g(x,y)=0$. Find the equations of these two fixed curves.

Answer 1

Re-write the given line $$y=\sin \theta (x-2)-9 \sin \theta +\sin 3\theta ~~~(0)$$ as $$y=s x -8s-4s^3~~~(1)$$ where $s=\sin \theta$ is a real parameter. Compare it with the normal of the parabola $y^2=4ax$ which is $$y=mx-2am-am^3.$$ This gives $a=4$, so the given family of lines (0) or (1) are normals of the parabola $f(x,y)=y^2-16x=0.$ Usually, a normal at one point of the curve cuts it at some other point also.

Next we set up the first order ODE for (1) eliminating $s$ by differentiating (1) w.r.t. $x$, we get $$s=y'~~~(2).$$ So the required ODE for the family of lines (1) is $$y=xy'-8y'-4y'~^3~~~(3)$$ This is the well known Clairaut's equation https://en.wikipedia.org/wiki/Clairaut%27s_equation, an ODE that has two solutions. One of the solutions is (1) where $s$ is a real parameter. The other one is a singular solution without a parameter. To get this singular (fixed) solution of (3), differentiate (3) w.r.t. $x$ to write $$y''(x-8-12y'~^2))=0~~~(4)$$ $$ \Rightarrow y''=0 ~ \mbox {or}~ y'~^2=\frac{x-8}{12} ~~~(5).$$ While $y''=0$, gives back (2) and hence the general solution of (3), where the parameter $s$ is the single constant of integration as the ODE (3) is first order. The second part of (5) gives the fixed singular solution as $$ 27 y^2=(x-8)^3,~ x\ge 8~~~(6).$$ Finally, Eq. (6) gives us the required fixed curve $g(x,y)=27y^2-(x-8)^3=0$, which touches the given family of lines (0) or (1). Note that (6) is a non-quadratic and non-conic curve whose tangent at one point may cut this curve at some other point.