Finding a cubic where tangent line at one point is normal at another intersection I have a cubic polynomial in terms of an unknown value $a>0$ that allows the tangent line at $x=1$ to also be the normal to the curve at $x=6$. The equation is given as $$g(x)=ax^2(x-8)$$
I have found the derivative of the equation to be $g'(x)=3ax^2-16a$, and the equation of the tangent at $(1,-7a)$ is $y=6a-13ax$. This tangent line intersects $g(x)$ at $(1,-7a)$ and $(6,-72a)$. 
I have tried equating the gradient of the tangent ($-13a$) with the gradient of the normal ($\frac{-1}{12a}$) which only results in $a$ being equal to $\frac{1}{156}$, which is way too small.
I have also tried finding equations for both the tangent and normal using the point-gradient formula: $y+y_1=m(x-x_1)$. For the tangent at $(1,-7a)$ with gradient $-13a$, $y=6a-3ax$; and the normal at $(6,-72a)$ with gradient $12a$: $y=\frac{-x+6}{12a}-72a$. Equating them at point $x=6$ gives $a=0$.
I have been stuck on this problem for a while now. Everything I try seems to yield the wrong answer. Thanks for any help.
 A: The slope of the tangent is $-13a$. Equate it to the slope of the normal, we know that
$$-13a=-\frac1{3a(36)-16a(6)}$$
$$6(13a)(18a-16a)=1$$
$$6(13a)(2a)=1$$
$$a^2 = \frac{1}{156}$$
$$a=\frac1{\sqrt{156}}$$

A: 
Find the value of $a$ so that the tangent to $y=g(x)=ax^2(x-8)$ at $(1,g(1))$ is also normal to the curve at the second point of intersection of the tangent with the curve. 

The tangent line to a function $g$ at $(1,g(1))$ is given by: $$y-g(1)=g'(1)(x-1)$$ $$y-(-7a)=-13a(x-1)\iff y=-13ax+6a=a(-13x+6)$$
Hence the tangent line is given by $y=-13ax+6a$. 
The point(s) of intersection between the tangent line and the curve is given by the solutions to the equation $$ax^2(x-8)=a(-13x+6)\iff x^3-8x^2=-13x+6\iff x^3-8x^2+13x-6=0\iff$$ $$(x-1)^2(x-6)=0\implies x=1, x=6$$ So the second point of intersection that we found between the tangent and curve is $$(6,g(6))$$
The tangent line of the curve at this point has slope $g'(6)=3a(6)^2-16a(6)=12a$
Remember: Any given line is normal to a given curve at a point $x_0$ if the tangent line at $x_0$ is perpendicular to the line at $x_0$. In other words, if the product of the tangent line's slope and the arbitrary line's slope is $-1$, then the line is normal to the curve at that point $x_0$.
For the tangent at $(1,g(1))$ to be normal to $g$ at the second point of intersection, we need $$(12a)(-13a)=-1\implies a^2=\frac{1}{156}\implies a=\frac{1}{\sqrt{156}}$$
A: Looks like you forgot to take a square root or lost a power of two along the way. The equation $-13a=-\frac1{12a}$ is equivalent to $a^2=\frac1{156}$, so the solution is $a=\frac1{\sqrt{156}}$, not $a=\frac1{156}$.
