MAT 2021 Question I was wondering about the following question taken from the MAT 2021:

The tangent to the curve $y = x^3 − 3x$ at the point $(a, a^3 − 3a)$ also passes through
the point $(2, 0)$ for precisely $$\text {(a) no values of $a$},\\ \text {(b) one value of $a$},\\ \text {(c) two values of $a$},\\ \text {(d) three values of $a$},\\ \text {(e) all values of $a$}$$

This is what is shown in the solutions:
The tangent at $a$ is $$y = (3a^2 − 3)(x − a) + (a^3 − 3a)$$ which passes through $(2, 0)$ if and only if
$$0 = (3a^2 − 3)(2 − a) + (a^3 − 3a)$$ This simplifies to $$2a^3 − 6a^2 + 6 = 0$$. The left-hand side is a cubic in $a$ and we’d like to know how many roots it has.
The turning points of $2a^3 − 6a^2 + 6$ are at $a = 0$ and $a = 2$, where the value of the cubic is $6$ and $−2$ respectively. So this cubic starts negative, rises to a positive local maximum, then decreases to a negative local minimum before rising again. There are therefore three roots for this cubic, so three values of $a$ for which the tangent to the original cubic passes through the point $(2, 0)$.
I understand it a little, but I really do not understand how they got the tangent to the equation.
As far as I understand, the derivative of the curve at a is $3a^2 -3$, and since it passed through the point $(2,0)$, we get the equation as $y-0 = (3a^2 - 3)(x - 2)$. This then simplifies to $y=(3a^2-3)(x-2)$, so implying that it intersects the point $(2,0)$ for all values of $a$.
I think I'm missing something very obvious here. Can someone please explain the solution as well as shed light on what I'm missing please?
 A: You said you didn’t understand how they got the equation of the tangent, so here’s the standard method.
$$y= x^3 - 3x$$
We want the gradient of the tangent at the point when $x=a$. This is done by taking the derivative.
$$\frac{dy}{dx}=3x^2-3$$
And so the gradient when $x=a$ is, $3a^2 - 3$. This tangent also passes through the point $(a,a^3-3a)$ so using the formula for the equation of a line given the gradient and a point, we get the tangent equation is:
$$y - (a^3 - 3a) = (3a^2 - 3)(x - a)$$
Then we substitute in $x=2$ and $y=0$ to get the cubic in $a$.
Where you went wrong is you found the gradient at the point where $x=a$ correctly, but then you set it to go through the point $(2,0)$. This isn’t actually a tangent to the curve in every case, it’s just a line with the same gradient as the tangent at $x=a$ which passes through $(2,0)$. If you plot such an equation on a graph, you’d see that your equation is only actually a tangent to the curve for 3 values of $a$, which would give you the 3 solutions you wanted.
A: 

The tangent to the curve $y = x^3 − 3x$ at the point $(a, a^3 − 3a)$ also passes through $(2, 0)$ for precisely $$\text {(a) no values of $a$},\\ \text {(b) one value of $a$},\\\cdots$$

the derivative of the curve at a is $3a^2 -3,$ and since the tangent passes through the point $(2,0)$, we get its equation as $y-0 = (3a^2 - 3)(x - 2).$

Ah, I see how your confusion arose. Observe that $a$ here is an arbitrary constant, and that we are investigating the effects of varying it. As $a$ varies, the tangent at the point $x{=}a$ may or may not pass through $(2,0),$ and we want to determine how many times this happens.

implying that it intersects the point $(2,0)$ for all values of $a$.

Your argument is circular, because your tangent equation was constructed under the assumption that the tangent line at the arbitrary point $x{=}a$ passes through $(2,0);$ in this case, of course $(2,0)$ lies on the tangent line for every value of $a.$
Instead, when constructing the tangent equation, just as you wrote the derivative $3a^2 -3$ in terms of $a,$ you should similarly choose a general point on the tangent line, that is, choose a point on the tangent line also in terms of $a,$ so, choose the given point $(a, a^3 − 3a)$ on the curve.
(Even when the point $(2,0)$ lies on the tangent line, it is not of the form $(a,a^3 -3a)$ as it does not lie on the curve.)
