Background
I am trying to solve the following problem:
> Given 2 distinct curves $C_1: y=f(x)=e^{6x}$ and $C_2: y=g(x)=ax^2$ where $a>0$. The objective is to find the range of $a$ such that there exists 2 tangents, each is tangent to both given curves.
I have solved it as follows.
Let $(t,e^{6t})$ be "a single point" on $C_1$ through which a tangent passes.
The slope of the tangent is $6e^{6t}$ that is obtained from the first derivative of $f(x)=e^{6x}$ at $x=t$.
The tangent is $$ y-e^{6t} = 6e^{6t}(x-t) $$
Let $(u,au^2)$ be a point on $C_2$ through which the tangent passes. It means that the tangent and $C_2$ have only one intersection point $(u,au^2)$.
We have, $$ au^2 -e^{6t} = 6e^{6t}(u-t) $$ By arranging it, we have a quadratic equation in $u$. $$ au^2 -6e^{6t}u + e^{6t}(6t-1) =0 $$
In order to make the quadratic equation have twin roots, its discriminant must be equal to zero as follows. $$ (-6e^{6t})^2-4\times a\times e^{6t}(6t-1) =0 $$
There will be 2 tangents if there are 2 roots $t_1$ and $t_2$ for the equation in $t$ below. $$ (-6e^{6t})^2-4\times a\times e^{6t}(6t-1) =0 $$
We can interpret finding the roots of the equation as finding the abscissa $t$ of the intersection of $\alpha(t)=a$ and $\beta(t)=\frac{9e^{6t}}{6t-1}$.
Using derivative, we can determine the minimum value of $\beta$ which is $9e^2$. Thus $a>9e^2$ is the required range.
My confusion
In the first step I must assume there is a single $t$ but in the last step I have to assume there are 2 roots $t_1$ and $t_2$. It looks inconsisten, doesn't it? What is wrong with my understanding?
I will repeat your solution with some justifications.
The tangent line to $y = e^{6 x}$ at the point $(\xi, e^{6\xi})$ is \begin{equation*} y = 6 e^{6\xi}(x-\xi) + e^{6\xi}. \end{equation*} A line in the above family is tangent to the graph $y = a x^2$ if and only if the equation \begin{equation*} a x^2 = 6 e^{6\xi}(x-\xi) + e^{6\xi} \end{equation*} has a unique solution for $x$. The last equation has a unique solution for $x$ if and only if
\begin{equation*} 9 e^{6\xi} -6 a \xi +a = 0, \end{equation*} that is, if and only if \begin{equation*} a = \frac{9 e^{6\xi}}{6 \xi -1} \end{equation*} Since $a \gt 0$, we conclude that the tangent line \begin{equation*} y = 6 e^{6\xi}(x-\xi) + e^{6\xi}. \end{equation*} is a tangent line to $y = a x^2$ if and only if \begin{equation*} a = \frac{9 e^{6\xi}}{6 \xi -1} \quad \text{and} \quad \xi \gt \frac{1}{6}. \end{equation*} There will be two tangent lines to $y = e^{6x}$ that are tangent to $y=ax^2$ if and only if the equation \begin{equation*} a = \frac{9 e^{6\xi}}{6 \xi -1} \end{equation*} has two solutions for $\xi$ such that $\xi > 1/6$. Finally, as you pointed out, this is the case if and only if $a \gt 9e^2$.