Two lines through a point, tangent to a curve We are looking for two lines through $(2,8)$ tangent to $y=x^3$. Let's denote the intersection point as $(a, a^3)$ and use the slope equation together with the derivative to get $\frac{a^3-8}{a-2}=3a^2$. This yields a cubic equation. Of course, one of the lines is tangent to $y=x^3$ at $(2,8)$, so we can get $a=2$ and the first line almost immediately. Knowing this, we write the cubic equation as $(a-2)(a^2+pa+q)$, find $p$ and $q$ from the original cubic equation and get $(a-2)(a-2)(a+1)=0$, which gives the solution and the lines. Is there a quicker way?
 A: Your work is correct. A bit more quickly you can write the equation that gives the points $(a,f(a))$ of tangency of a line passing through a point $P=(x_P,y_P)$ as:
$$
f(a)-y_P=f'(a)(a-x_P)
$$
that, in this case becomes:
$$
a^3-8=3a^2(a-2)
$$
so $a=2$ is a trivial solution and the other solutions are given by:
$$
a^2-2a+4=3a^2 \quad \iff \quad (a-2)(a-1)=0
$$
Note that the solution$a=2$ is a double solution of the inital equation because the point $P$ is a point of the given function.
A: The tangents must be of the form
$$y-8=m(x-2),$$ and they intersect the cubic $\color{blue}{y=x^3}$ when
$$x^3-8=m(x-2).$$
This equation must have a double root, so that differentiating on $x$, we also have
$$3x^2=m.$$
With the obvious solution $x=2$, we deduce $m=12$ and
$$\color{green}{y-8=12(x-2)}.$$
Otherwise, we may simplify to get
$$x^2+2x+4=m=3x^2.$$
This gives another solution $x=-1$, then $m=1$ and
$$\color{magenta}{y-8=3(x-2)}.$$

A: There's another way, not sure it's quicker in the present case.
Consider a line with slope $t$ passing through the point $(2,8)$. This line will be tangent to the curve if the abscissae equation for intersection points has a root  of multiplicity $>1$.
Now the equation of such a line is $\;y=t(x-2)+8$, so the abscissae equation is
\begin{align*}
(x-2)+8=x^2=3&\iff x^3-8=(x-2)(x^2+2x+4)=t(x-2)\\
&\iff\begin{cases}x=2\\x^2+2x+4=t
\end{cases}
\end{align*}
The quadratic equation $x^2+2x+4=t$ has a double root at its minimum, obtained for $x=-1$, and this minimum is $t=3$.
So the two lines are


*

*the tangent at $(2,8)$, with equation $\;y=12(x-2)+8=12x-16$,

*the tangent at $(-1,-1)$, with equation $\;y=3(x-2)+8=3x+2$. Alternatively, this equation is also $\;y=3(x+1)-1$.

