Find point at constant distance from parabola

Question

Kindly consider the diagram below, with:

• f(x) = x²
• x₀ given.
• d constant, e.g. 1 in the diagram.
• y₀ unknown.

I would like to calculate point (x₀, y₀), sitting perpendicular to the curve f(x), at a constant distance of d, given x₀. How do I best approach this problem?

To be clear, I do not know x₁ in advance. I've merely labeled it for convenience.

Answer 1

I'll work with the parabola $y=x^2$ and the distance $d=1$ for convenience' sake, but the method should be tweakable easily enough.

The tangent vector to the parabola at $x=x_1$ has slope $2x_1$; that is to say, it points in the (unnormalized) direction $\langle 1, 2x_1\rangle$. The orthogonal vector to this (pointing downwards, as in the diagram) then points in the direction $\langle 2x_1, -1\rangle$. The length of this vector is $\sqrt{1+4x_1^2}$, so the normalized vector is $\langle \frac{2x_1}{\sqrt{1+4x_1^2}}, \frac{-1}{\sqrt{1+4x_1^2}}\rangle$, of unit length. The point at distance one from the parabola along this orthogonal line at $\langle x_1, y_1\rangle = \langle x_1, x_1^2\rangle$ is then $\langle x_1+\frac{2x_1}{\sqrt{1+4x_1^2}}, x_1^2-\frac{1}{\sqrt{1+4x_1^2}}\rangle$. We want the $x$ component of this equal to $x_0$, so we set $x_0=x_1+\frac{2x_1}{\sqrt{1+4x_1^2}}$ and solve for $x_1$: $x_0-x_1 = \frac{2x_1}{\sqrt{1+4x_1^2}}$; $(x_0-x_1)^2=\frac{4x_1^2}{1+4x_1^2}$; $(1+4x_1^2)(x_0-x_1)^2-4x_1^2=0$. This is a quartic equation in $x_1$; it can be solved explicitly, but for practical purposes iterative root-finding is likely to yield better results.