Some years ago I thought of a problem that I've returned to frequently but never been able to solve. Maybe it's impossible, I'm not a mathematician. I thought I should ask some experts rather than puzzling over it till my dying day. The problem involves a simple looking curve constructed from the parabola $y = x^2$ as follows. From the point $(x , x^2)$ construct a normal to the curve (in the direction of increasing $y$) which is one half unit long. The end point of this normal segment traces out a new curve $Y = F (X)$ as $x$ varies. The problem is to express $Y$ explicitly in terms of $X$.
From the diagram -
$tan{\ \theta} = 2x = \dfrac{dY}{dX} = \dfrac{x - X}{Y - x^2}$
also ${({\dfrac{1}{2})}^2} = {{(x - X)}^2} + {{(Y-x^2)}^2}$
Considering this I could express $X, Y$ and $Y'$ parametrically in terms of $x$ -
$X = x - \dfrac{x}{\sqrt{1 + 4 x^2}}$
$Y = x^2 + \dfrac{1}{2\sqrt{1 + 4 x^2}}$
$\dfrac{dY}{dX} = 2x$
I've never got any further than this. I've never been able to construct a solvable looking differential equation or polynomial or eliminate the parameter. After all these years it's got to stop. If it's not possible, can one demonstrate that it is not possible? What kind of curve is this? Any assistance will be gratefully received.
The problem is essentially to eliminate the parameter $t$ from the pair of equations $$x=t-\frac{t}{\sqrt{1+4t^2}}$$ $$y=t^2+\frac{1}{2\sqrt{1+4t^2}}$$
To do this, let $$2t=\sinh u\implies \sqrt{1+4t^2}=\cosh u$$
Then, $$x=\frac12(\sinh u-\tanh u)$$ and $$y=\frac14\sinh^2u+\frac{1}{2\cosh u}=\frac14\left(\cosh^2u-1+\frac{2}{\cosh u}\right)$$
Meanwhile, after some simplification, we get $$x^2=\frac14\left(\cosh^2u-2\cosh u+\frac{2}{\cosh u}-\frac{1}{\cosh^2u}\right)$$
So now we can obtain $$4y-4x^2+1=2\cosh u+\frac{1}{\cosh^2u}=:\lambda$$ and also $$4y+1=\cosh^2u+\frac{2}{\cosh u}=:\mu$$
From the second of these equations we can get $$\cosh^3u=\mu\cosh u-2$$
Substituting this expression for $\cosh^3u$ into the first equation gives $$\lambda=\frac{2(\mu\cosh u-2)+1}{\cosh^2u}\implies\lambda\cosh^2u-2\mu\cosh u+3=0$$
Hence, $$\cosh u=\frac{\mu\pm\sqrt{\mu^2-3\lambda}}{\lambda}$$
The $\pm$ indicates the fact that there are two options for the locus: one above the parabola and one below.
From now onwards it gets very messy to actually obtain the Cartesian equation(s). First we have to get an equation just in $\lambda$ and $\mu$ and then substitute into that their expressions in terms of $x$ and $y$.
Firstly, $$\cosh^2u=\frac{2\mu^2-3\lambda\pm2\mu\sqrt{\mu^2-3\lambda}}{\lambda^2}$$
Therefore the equation in terms of $\lambda$ and $\mu$ is $$\mu=\frac{2\mu^2-3\lambda\pm2\mu\sqrt{\mu^2-3\lambda}}{\lambda^2}+\frac{2\lambda}{\mu\pm\sqrt{\mu^2-3\lambda}}$$
I had a go at rewriting this without the square roots, and I got$$(\mu^2\lambda^2-4\mu^3-2\lambda^3+9\lambda\mu)^2=\pm(\mu^2-3\lambda)(\mu\lambda^2-4\mu^2+3\lambda)$$
You may want to check this for yourself, but in any event you now have to substitute into this equation (or the previous one) the expressions $\mu=4y+1$ and $\lambda=4y-4x^2+1$ to get the Cartesian equation(s).
I will leave this to you. I hope this helps.