For all $h\in \mathbb{R}\sim\{0\}$, two distinct tangents be drawn from the point $(2+h,3h-1)$ to the curve $y=x^3-6x^2-a+bx$ then $\frac{a}{b}=?$

Question

If for all $h\in \mathbb{R}\sim\{0\}$, two distinct tangents can be drawn from the points $(2+h,3h-1)$ to the curve

$y=x^3-6x^2-a+bx$

then find value of $\frac{a}{b}$

My Attempt

If two distinct tangents are to be drawn from point $P\equiv(2+h,3h-1)\forall h\in\mathbb{R}\sim \{0\}$, then $P$ can be any point on the line $y=3x-7$ except $(2,-1)$.

By plotting the graph and visualizing in all possible ways I came to the conclusion that one of the tangents must be the tangent line at the point of inflexion and that tangent line must be the line $y=3x-7$.

So, I proceeded further

$y'(x)=3x^2-12x+b$ and $y''(x)=6(x-2)$. So we have point of inflection at $x=2$.

Consequently, $y'(2)=b-12$ and $y(2)=2b-a-16$.

So equation of tangent line through the point of inflection is

$y=y(2)+y'(2)(x-2)$

i.e. $y=(b-12)x+8-a$

If this were to coincide with $y=3x-7$ then $b-12=3$ and $8-a=-7$ which give $a=b=15$ and so $\frac{a}{b}=1$ which is indeed the right answer.

Another observation is that now the equation of the curve is

$y=x^3-6x^2+15x-15$ and the point $(2,-1)$ lies on the curve

But how to explain this idea analytically or mathematically. Visualization would not count in an exam.

Answer 1

Let $f(x)=x^3-6x^2+bx-a.$

The discriminant or resultant of the expression and its derivative wrt $t$ of $$y-f(t)-(f'(t)(x-t))$$ or

3*h-1-(t^3-6*t^2+b*t-a)-((3*t^2-12*t+b)*(h+2-t))

is given by the determinant of the five by five Sylvester matrix

$\begin{vmatrix}2&-3\,h-12&12\,h+24&-b\,h+a-2\,b+3\,h-1&0\\ 0&2&-3\,h-12&12\,h+24&-b\,h+a-2\,b+3\,h-1\\ 6&-6\,h-24&12\,h+24&0&0\\ 0&6&-6\,h-24&12\,h+24&0\\ 0&0&6&-6\,h-24&12\,h+24 \end{vmatrix}.$

Maybe not tractable by hand, but a CAS simplifies this to

216*(b*h-15*h+2*b-a-15)*(h^3+b*h-15*h+2*b-a-15)

The linear factor vanishes identically when $a=b=15$. With that the second factor reduces to $h^3$.