How to find the envelope of the family of curves $(x-t)^2 + (y-t^2)^2 = 1$? How to find the envelope of the family of curves $(x-t)^2 + (y-t^2)^2 = 1$?
Based on the given, the family of curves can be describe a set of circles whose centers are the points along the parabola $x=y^2$.
To find the envelope, I am aware that I have to find the partial derivative of the given with respect to $t$ and solve simultaneuously the given equation and its partial derivative with respect to $t$.
Here is my incomplete solution.
Let $f(x,y,t)=(x-t)^2 + (y-t^2)^2 - 1=0$
$f_t(x,y,t)=-2(x-t) + (-4t)(y-t^2)=0$
$f_t(x,y,t)= (x-t) + (2t)(y-t^2)=0$
What is the most efficient way to solve simultaneously $f$ and $f_t$?
Please feel free to share your thoughts. Any comment and suggestion will be highly appreciated.
 A: What you have is the general way to find the envelope of a family of curves. It doesn't look very easy, and I think there are easier ways to do this.
You want the values of $x$ and $y$ for which $f$ has a repeated root with respect to $t$. That is what it means that a function and its derivative have a common root. Luckily, this is a solved problem for polynomials, and it is given by where the discriminant is equal to $0$
Unfortunately, however, the answer is a little complicated:
$$
256 x^6 + 256 x^4 y^2 - 640 x^4 y - 752 x^4 - 512 x^2 y^3 + 96 x^2 y + 448 x^2 + 256 y^4 - 640 y^3 + 144 y^2 + 640 y - 400=0
$$
I don't think I would personally have had the perseverance to extract that from your two equations, and I'm happy that I can just look up the fourth degree discriminant, or in this case make WolframAlpha look it up for me.
A: Parametric equation of parabola is $p(t) =(x,y) = (t, t^2) $
Unit tangent vector is $ u = \dfrac{ (1, 2 t) }{\sqrt{1 + 4 t^2}} $
Unit normal vector is $ v = \dfrac{ (-2 t, 1 ) }{\sqrt{1 + 4 t^2}}$
Normal line at $t$  is $q(t) = (t, t^2) + s \dfrac{(-2 t, 1 )}{\sqrt{1 + 4 t^2}}$
Take $s = 1$ (because the radius of the circle is $1$ ) then the inner envelope is parametrically given by
$q(t) = ( t - \dfrac{2 t}{ \sqrt{1 + 4 t^2}} , t^2 + \dfrac{1}{\sqrt{1 + 4 t^2}} )$
or you can take s = -1 and this lead to the outer envelope.
$q(t) = ( t + \dfrac{2 t}{ \sqrt{1 + 4 t^2}} , t^2 - \dfrac{1}{\sqrt{1 + 4 t^2}} )$
