# Determining the points at constant distance $d$ from parabola $y = x^2$

My friends recently learned about Locus and how parabola are the points equdistant from a fixed point (focus) and a straight line (directrix). However, now we're trying to find an equation or set of equations for the points at a constant distance from the parabola of function y = x^2 and cannot be closer/further from any point on the curve than any arbitrary distance (just use variable d).

As they have more experience with derivatives, they've gone with using the equation of the normal to find these equations. I've tried to use the distance formula, but with no success.

It would be awesome if we got some help on this!

• I think you'll need to find normal at each point on given parabola. Then point at distance $d$ along the normal gives corresponding point on new curve – samjoe May 25 '18 at 7:32
• Welcome to MSE. It is in your best interest that you use MathJax. – José Carlos Santos May 25 '18 at 7:34
• So you want a solution without using derivatives? What is the distance formula? – mvw May 25 '18 at 8:31

We have $f(x) = x^2$. A tangent vector at each point $P(x)= (x, f(x))$ is $$t(x)=(1, f'(x))$$ The normal vectors are orthogonal to the tangent vectors which gives $$n_i(x) = (\mp f'(x), \pm 1)$$ and satisfies $$t(x) \cdot n_i(x) = 1 \cdot (\mp f‘(x)k + f‘(x) \cdot (\pm 1) = 0$$ As we want to specify a distance along a normal, it is easier to use normal vectors with length $1$ (unit vectors): $$n_i(x) = \frac{(\mp f'(x), \pm 1)}{\sqrt{1+f'(x)^2}}$$ The two cases of sign stand for the normal vector pointing inwards and outwards. Then a point $Q(x)$ with distance $d$ along a normal to $P(x)$ is $$Q(x) = P(x) + d \, n_i(x) = (x, f(x)) + d \, \frac{(\mp f'(x), \pm 1)}{\sqrt{1+f'(x)^2}}$$ In this case $$Q(x) = (x, x^2) + d \frac{(\mp 2x, \pm 1)}{\sqrt{1+4x^2}}$$