Single Variable conics.

Let $f(x)=x^3-6x^2-a+bx$ ; we will first determine the equation of the straight line tangent to $f(x)$ , whose gradient is a minimum. Letting $f''(x)=0$ , we obtain $6x-12=0\Longrightarrow x=2$ , and hence the line has gradient $f'(2)=b-12$ . Substituting the point $(2,f(2))$ allows us to determine the $c$ -value, and thus the equation to be $y_T=(b-12)x+8-a$ . Now for $h>0$ , note that

$(b-12)(2+h)+8-a<3h-1<(2+h)^3-6(2+h)^2-a+b(2+h)$

for there to be two distinct tangents to $f(x)$ from $(2+h,3h-1)$ , due to the concavity of $f(x)$ for points $(2+h,3h-1)$ where $3h-1<(b-12)(2+h)+8-a$ or $3h-1>(2+h)^3-6(2+h)^2-a+b(2+h)$ .

Similarly, note that for $h<0$ ,

$(b-12)(2+h)+8-a>3h-1>(2+h)^3-6(2+h)^2-a+b(2+h)$

for there to be two distinct tangents to $f(x)$ from $(2+h,3h-1)$ , due to the concavity of $f(x)$ for points $(2+h,3h-1)$ where $3h-1>(b-12)(2+h)+8-a$ or $3h-1<(2+h)^3-6(2+h)^2-a+b(2+h)$ .

Hence, we determine that any point $(2+h,3h-1)$ must lie in one of two disjoint regions between $f(x)$ and $y_T=(b-12)x+8-a$ . Note that the fact that the requirement must be satisfied for all values of $h$ (excluding $0$ ) is equivalent to implying the existence of the line $y=3x-7\;\; \forall \;\; x\setminus \{2\}$ lying strictly within these regions. It is then clear that this line must be the tangent line; this is due to the existence of a solution to $f(x)=3x+7,\; x\neq 2$ for all values of $a$ and $b$ , except when the only point of intersection is at $x=2$ , wherein it is not in fact an intersection since $x\neq 2$ . The only line that intersects $f(x)$ at $x=2$ only, and whose gradient is no less than that of $y_T$ (so that it lies in the two regions) is clearly $y_T$ itself. Hence, we conclude $y_T=(b-12)x+8-a=3x-7\;\; \forall \;\; x\in \mathbb{R}$ .

Hence, $(b-12)(2+h)+8-a=3h-1$ for all values of $h$ , meaning that $a=b=15$ is the only possible solution for $(a,b)$ . Thus $\dfrac{a}{b}=\boxed{1}$ .

A line through $P\;(2+h,3h-1)$ will be of the form $y \; = \; m(x-h-2) + 3h-1$ for some constant $m$ , and it will be a tangent to the cubic provided that the cubic equation $f(x) \; = \; x^3 - 6x^2 - a + b x - m(x - h - 2) - (3h - 1) \; = \; 0$ has a repeated root, which means that the simultaneous equations $f(x) = f'(x) = 0$ have a root. Since we can write $f(x) \; = \; A(x)f'(x) + B(x) \hspace{2cm} f'(x) \; = \; C(x)B(x) + D$ where $A(x), B(x), C(x)$ are linear polynomials and $D$ is a constant, we deduce that we must have $D=0$ at a tangent. We calculate that $D \; =\; \frac{g(m)}{4(12 - b + m)^2}$ where $g(m)$ is a cubic polynomial in $m$ (with coefficients in $a,b,h$ ). The exact form of $g(m)$ is too complicated to type out easily!

Thus we have a tangent to the original cubic passing through the point $P$ with gradient $m$ provided that $g(m) = 0$ . Since we want there to be exactly two tangents through $P$ for each nonzero $h$ , and since $g(m)$ is a cubic polynomial, we deduce that the cubic $g(m)$ must have a repeated root. Playing the same quotient/remainder game as above, since we can write $g(m) \; = \; \alpha(m)g'(m) + \beta(m) \hspace{2cm} g'(m) \; = \; \gamma(m)\beta(m) + \delta$ where $\alpha(m), \beta(m), \gamma(m)$ are linear polynomials in $m$ and $\delta$ does not depend on $m$ , we deduce that we must have $\delta = 0$ . Now $\delta \; = \; -\frac{432\big[(15 + a - 2b) + (15-b)h - h^3\big]\big[(15 + a - 2b) + (15-b)h\big]^3}{h^2\big[8(15+a-2b) + 8(15-b)h - 9h^3\big]^2}$ and the only way we can ensure that $\delta = 0$ for all nonzero $h$ is to have $15+a-2b = 15-b = 0$ , so to have $a=b=15$ . Thus $\tfrac{a}{b} = \boxed{1}$ .