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

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

Google Colab