Finding a parametric curve The parabola y=x² is parameterized by x(t) = t and y(t) = t². At the point A (t,t²) a line segment AP 1 unit long is drawn normal to the parabola extending inward. Find the parametric equations of the curve traced by the point P as A moves along the parabola.
Best picture i could come up with
 A: According to your picture you're on the right track.  Add to each point $(t,t^2)$ the normalized normal vector $(-2t,1)/\sqrt{4t^2+1}$ so that the equation of the parallel curve is given by
$$(t,t^2)+(-2t,1)/\sqrt{4t^2+1}=\left(t-\frac{2t}{\sqrt{4t^2+1}},
t^2+\frac{1}{\sqrt{4t^2+1}}\right).$$
A hard question without having been teached to deal with vectors.
A: The parabola can be parametrised by $\gamma(t)=(t,t^2)$.
The unit tangent vector is given by $${\bf T} = \frac{\dot{\gamma}}{\|\dot\gamma\|}=\left( \frac{1}{\sqrt{1+4t^2}}, \frac{2t}{\sqrt{1+4t^2}}\right)$$
The unit normal vector is given by rotating this anti-clockwise by $90^{\circ}$:
$${\bf N} = \left(\frac{-2t}{\sqrt{1+4t^2}} , \frac{1}{\sqrt{1+4t^2}}\right)$$
The curve you're interested in is an example of a wave front. They are given by $\gamma + \lambda {\bf N}$
$$\gamma + \lambda {\bf N} = \left(t-\frac{2\lambda t}{\sqrt{1+4t^2}} , t^2+\frac{\lambda}{\sqrt{1+4t^2}}\right)$$
for different values of $\lambda$. In your case, you have $\lambda  = 1$.
An interesting result says that a wave front is singular at $\gamma(t_0)+\lambda{\bf N}(t_0)$ when $\lambda = 1/\kappa(t_0)$, where $\kappa$ is the curvature of $\gamma$, i.e. when $\lambda$ is a radius of curvature.
