$$\newcommand{\A}[1]{\colorbox{lightgray}{\displaystyle #1}} \newcommand{\B}[1]{\colorbox{CornflowerBlue}{\displaystyle #1}} \newcommand{\C}[1]{\colorbox{BurntOrange}{\displaystyle #1}}$$

Let

$f(x)=ax^2$

Let

$A=(x_A,y_A)$

be a point on the parabola $$f$$. Then

$y_A=f(x_A),$

i.e.

$y_A=ax_A^2.$

Let

$f_m(x)=m(x-x_A)+y_A$

and let

\begin{align*} g_m(x)&=f(x)-f_m(x)\\ g_m(x)&=ax^2-\left(m(x-x_A)+y_A\right)\\ g_m(x)&=ax^2-mx+mx_A-y_A \end{align*}

Let $$D_m$$ be the discriminant of $$g_m(x)$$:

\begin{align*} D_m &= (-m)^2 - 4(a)(mx_A-y_A)\\ D_m &= (-m)^2 - 4(a)(mx_A-ax_A^2)\\ D_m&=m^2 -4amx_A+4a^2x_A^2\\ D_m&=(m-2ax_A)^2. \end{align*}

On the one hand, $$D_m=0$$ is equivalent to $$m=2ax_A$$.

On the other hand, $$D_m=0$$ is equivalent to the line $$f_m$$ and the parabola $$f$$ having intersection multiplicity 2 at the point $$A=(x_A,y_A)$$.

Let

$m_A=2ax_A$

and let

\begin{align*} f_A(x)&=f_{m_A}(x)\\ f_A(x)&=m_A(x-x_A)+y_A\\ f_A(x)&=2ax_A(x-x_A)+y_A. \end{align*}

Thus, for

$\A{m_A=2ax_A}$

the line

$\A{f_A(x)=m_A(x-x_A)+y_A}$

is tangent to the parabola $$f(x)=ax^2$$ at the point $$A=(x_A,y_A)$$.

Intersection of parabola and line