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?

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
$$\color{magenta}{y-8=3(x-2)}.$$

