Special Lines!

$\displaystyle f(x)=x^3+ax^2+bx+c$

$\displaystyle f'(x)=3x^2+2ax+b$

For convenience, let the point of tangency of a special line have coordinates $(t,f(t))$ and the point of intersection, where the special line is normal to the curve, have coordinates $(n,f(n))$ .

We have:

$\displaystyle f'(t)\cdot f'(n)=-1$

$\displaystyle \frac{f(t)-f(n)}{t-n}=f'(t)$

Second equation leads to $\displaystyle (n-t)(n+a+2t)=0\implies n=-a-2t$ .

Substituting into the first and using $m=f'(x)$ , this yields $\displaystyle (3t^2+2at+b)\cdot(12t^2+8at+a^2+b)=-1\implies4m^2+(a^2-3b)m+1=0$ . Since the slopes of the two lines is the same, this quadratic equation must only have one root, implying $\displaystyle (a^2-3b)^2=16\implies 4m^2\pm 4m+1=0\implies m=\mp\frac{1}{2}$ . However, $\displaystyle m=\frac{1}{2}$ leads to a non-real $t$ .

Let $f(x) = x^3 + ax^2 + bx + c \implies f'(x_{0}) = 3x_{0}^2 + 2ax_{0} + b \implies$ the tangent line to the curve at $(x_{0},y_{0})$ is $y - (x_{0}^3 + ax_{0}^2 + bx_{0} + c) = (3x_{0}^2 + 2ax_{0} + b)(x - x_{0})$

Let the tangent line be normal to the curve at $(x_{1},y_{1}) \implies (x - x_{0})(x_{1}^2 + (a + x_{0})x_{1} - (2x_{0} + a)x_{0}) = 0$

$x_{1} \neq x_{0} \implies x_{1} = -(a + 2x_{0})$

Since the tangent is also normal to the curve at $(x_{1},y_{1}) \implies -1 =$ $(3x_{0}^2 + 2ax_{0} + b)(12x_{0}^2 + 8ax_{0} + a^2 + b) \implies 36x_{0}^4 + 48ax_{0}^3 + (19a^2 + 15b)x_{0}^2 + 2a(a^2 + 5b)x_{0} + (a^2 + b)b + 1 = 0$ .

Let $a = 0 \implies 36x_{0}^4 + 15bx_{0}^2 + b^2 + 1 = 0 \implies x_{0}^2 = \dfrac{-15b \pm 3\sqrt{9b^2 - 16}}{72}$ and letting $9b^2 - 16 = 0 \implies b = \pm\dfrac{4}{3}$ .

For real $x_{0}$ choose $b = -\dfrac{4}{3} \implies x_{0}^2 = \dfrac{5}{18} \implies x_{0} = \pm\dfrac{\sqrt{5}}{3\sqrt{2}} \implies$ $m_{1} = m_{2} = -\dfrac{1}{2}$ .

The cubic can be translated on the plane without affecting the slope to transform it to $f(x)=x^{3}-a^{2}x$ . With derivative $f'(x)=3x^{2}-a^{2}$

$(t,t^{3}-a^{2}t)$ is some point on the curve and the tangent line has equation $y-(t^{3}-a^{2}t)=(3t^{2}-a^{2})(x-t)$

Solve $f(x)=y$ to find where this line hits the cubic again: Besides $x=t$ the equation $x^{3}-a^{2}x=(3t^{2}-a^{2})x-2t^{3}$ has solution $x=-2t$

The slope of $f(x)$ at this point is $12t^{2}-a^{2}$

Now solve to find when the two slopes are perpendicular: $(12t^{2}-a^{2})(3t^{2}-a^{2})=-1$ whose only real solutions are

$a=\pm \frac{2}{\sqrt{3}}$ and $t= \pm \frac{\sqrt{10}}{6}$

Finally, the slope is then $3(\frac{\sqrt{10}}{6})^{2}-(\frac{2}{\sqrt{3}})^{2}=\boxed{-0.5}$

In retrospect using $a$ instead of $a^{2}$ in the original cubic would have been simpler, but I originally thought in terms of the x-intercepts being $(\pm a,0)$ .

This problem reminds me of an earlier problem on Sept-03, where the square of maximum area was placed on cubic curve.

Here too, assuming symmetry around the origin, the cubic curve is y=x^3-gx, where g is to be found within the given constraints of the tangent cutting the curve as a normal.

In the Sept-03 problem, g=√8, while here, it is g= 4/3, and after some algebra,

the slope of tangent/normal = -(1/2).