Find the function $g(x)$ that has a constant perpendicular distance to a known function $f(x)$ This is harder than I thought. Just trying to find the $g(x)$, a function that is always $a$ length units away from a known function $f(x)$, perpendicularly
I find myself having to mix linear algebra with calculus, which doesn’t work for my high school students. It feels like it should be totally possible, but I’m at my wit's end right now.
 A: do Carmo  calls the   construction a parallel curve.
Start with a parametrized curve $x(t), y(t)  .$     The vector $(\dot x , \dot y)$  is the tangent direction of the curve, and a perpendicular  direction is
$(- \dot y , \dot x).$   This is not necessarily constant length;  a unit normal direction is given by
$$\left(  \frac{ - \,\dot y}{\sqrt {\dot x^2  + \dot y^2}} , \frac{ \dot x}{\sqrt {\dot x^2  + \dot y^2}} \right).$$
We may multiply this by a desired constant distance, also a minus sign to go the other side of the curve.   Then add the position vector  and the chosen normal (multiple)  to get a new parametized curve. It may or may not be feasible to find an implicit relation the new curve satisfies. Begin with a circle, you get a concentric circle, and this will be visible in sine and cosine terms.   I think it is already a mess to do an ellipse.
Ellipse, $(3 \cos t, \sin t)$   for $x^2 + 9 y^2 = 9$
give me a few minutes
"Outward" unit normal is
$$\left(  \frac{ \cos t}{\sqrt {\cos^2 + 9 \sin^2 t}} ,
 \frac{ 3 \sin  t}{\sqrt {\cos^2 + 9 \sin^2 t}} \right).$$
add to position to get  new curve
$$  (x,y) = \left( 3\cos t  + \frac{ \cos t}{\sqrt {\cos^2 + 9 \sin^2 t}} ,  \sin t +
 \frac{ 3 \sin  t}{\sqrt {\cos^2 + 9 \sin^2 t}} \right).$$
Wikipedia says that this new curve is algebraic of degree eight. That is, take the expressions for $x,y$  and keep squaring things while consciousness lasts.  Deep breaths.
later: Viktor Vaughn  gave a link to sage   and an interactive diagram.   The first image has the ellipse in blue, the red curve inside by subtracting a quarter of the unit normal. Next subtract   one third, when the red curve first displays corners. Finally subtracting half the unit normal, obvious cusps.



A: Find out the equation of tangent of any given point (x1, f(x1))
Use the foot of perpendicular formula in the reverse sense, i.e. finding out the the point(h(x1), g(y1)) from the coordinates of the foot of the perpendicular (x1, y1) as opposed to how it is usually used in the opposite send and find out a relation between h(x1), g(y1), x1, f(x1)=y1, there'll be two independent equations and the other equation will be obtained by using distance formula and equating the distance from(h(x1),g(y1)) to (x1, f(x1)) with and these three equations should give you h(x) and g(y), It can be easily proved that it will be multivariable equation, as it can be easily visualized that there will be two points satisfying the three equations for one given point of the function which cannot be true for f(x) = y but can be true for (x2, y2) = (h(x1), g(y1))
Hope this helps, btw I think everyone will be able to understand this, since I'm a highschool student myself from India and it would be better to say that even though I have an interest in mathematics maths is NOT my forte
