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?

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 $\;y=3(x+1)-1$.

