How can you find the equation of the line that is tangent at two distinct points to the curve? Given a curve $y = x^3-x^4$, how can I find the equation of the line in the form $y=mx+b$ that is tangent to only two distinct points on the curve?
The problem given is part of the Madas Special Paper Set. This paper set, seems to not have any answers as Madas himself only released the answers on request. Sadly, Madas passed away this year and the contact information has been removed. So I am here to ask how to solve this question. I have tried to create some simultaneous equations but I cannot get to a resolute answer. I have also tried to differentiate and find some equation for the lines gradient but have come to no success: Please help! Question Here
 A: If you call $f(x)=x^3-x^4$, and you consider the function
$$
F(x_1,x_2)=\frac{f(x_1)-f(x_2)}{x_1-x_2}
$$
giving you the slope of the line joining two points on the graph, you can easily see that in the optimal situation (when the line is tangent), if you fix $x_1$, the function has a critical point w.r.t.$x_2$ and vice versa. In one case it is a local max in the other it is a local min. So basically we are trying to find a saddle point for $F$.
You have the optimality conditions
$$
\frac{\partial F}{\partial x_1}=2x_1+x_2-3x_1^2-2x_1x_2-x_2^2=0;
$$
$$
\frac{\partial F}{\partial x_2}=2x_2+x_1-3x_2^2-2x_1x_2-x_1^2=0.
$$
Disregarding the situation $x_1=x_2$ (corresponding to the degenerate case) and subtracting the above equations you find
$$
x_1+x_2=\frac 1{2}
$$
which, when plugged into the first optimality condition, gives a quadratic equation for $x_2$:
$$
\frac 1{4}+x_2-2x_2^2=0
$$
with solutions $x_2=(1\pm\sqrt{3})/4$
Taking into account the symmetry of your equations, you conclude that the points you are looking for have $x$-coordinates
$$
\frac{1+\sqrt{3}}{4};\qquad \frac{1-\sqrt{3}}{4} 
$$
I did the computations on a napkin, make sure that everything is correct.
A: If the graph of an unknown linear function $L(x)= ax+b$  grazes that of the given monic quartic $Q_4(x)= x^4 - x^3$ from below, as in your picture, then  their difference $Q_4(x)- L(x)$ vanishes to second order  at the two points of tangency, hence must have two double roots, thus factor as
(1) $Q_4(x)- L(x)= (Q_2(x)^2)$ where
(2) $Q_2(x) =(x-r_1)(x-r_2)$ is a monic quadratic. Note that the quadratic $Q_2(x)$ is symmetric (even) with respect to a line midway between its roots. Thus it can be written as
(3) $Q_2(x)= (x-m)^2 -c^2$ where $x=m$ is the midpoint.
Differentiate (1) twice to eliminate $L(x)$ and deduce
(4) $Q_4''(x) = 2 [Q_2(x) Q_2''(x)  + (Q_2')^2]$ and note that this quadratic is also even with respect to $x=m$.  (It is the sum of products of expressions that have this even symmetry.)
The left side of (4) is a known function of $x$, quadratic in $x$.  Explicitly, in this problem $Q_4''(X)= 12 x^2 - 6 x =x(12x-6)$.
Finding its line of symmetry (mid-point of roots) determines $m=1/4$. Once $m$ is known,  solve for $c$ by substituting (3) into (4) evaluated at $x=m=1/4$. Deduce $c=\pm \sqrt{3}/4$.
Note that this determines $Q_2(x)$ completely: $Q_2(x)= (x-\frac{1}{4})^2 -\frac{3}{16}$.
Additional response to OP's comment.
The line $L$ is the line that passes through the two points on the graph of the quartic that have as their $x$ coordinates the values $x_1=r_1= m-c$ and $x_2= r_2= m+c$. It can also be written simply as $L(x)= Q_4(x)- Q_2(x)^2$.
A: $(x,y)=(t,t^3-t^4),$ has dual $(X,Y)=(-\frac{y'}{xy'-x'y},\frac{x'}{xy'-x'y})=(-\frac{3t^2-4t^3}{t(3t^2-4t^3)-(t^3-t^4)},\frac1{t(3t^2-4t^3)-(t^3-t^4)})$ and implicitizes to $27X^4+4X^3Y-6X^2Y+192XY+27Y^2+256Y=0$ which has a node at $(8,-64)$ (and a cusp at $(-4,16)$) corresponding to the bitangent $8x-64y+1=0$ (and flex $-4x+16y+1=0$).
Edit
The dual curve lives in the dual projective plane of lines in the plane. It encodes all the tangent lines of the curve. A double point of this dual curve naturally encodes a bitangent for the original curve.
As for finding the dual, I've used the  formula for a parametrized curve. You could plot it and see the double point. I've elected to find a formula for it, and the system of the partial derivatives has solutions $(-4,16)$ (more precisely $\langle 8x+y+16,(y-16)^2\rangle$ ) and $(8,-64).$
Then the correspondence to get the lines goes through the equation for the universal line $xX+yY+1=0.$
