Problem source

Let $P$ be the point on  the curve $y=ax^2+bx+c$ (where $a$ is non-zero) at which the gradient is $m$. 
Show that the equation of the tangent at $P$ is 
\[ 
y-mx=c-\frac{(m-b)^2}{4a}\;. 
\] 
 
Show that the curves $y=a_1 x^2+b_1 x+c_1$ 
and $y=a_2 x^2+b_2 x+c_2$ (where $a_1$ and $a_2$ 
are non-zero) have a common tangent with gradient $m$ if and only if 
\[ 
(a_2 -a_1 )m^2 +
2(a_1 b_2-a_2 b_1)m +
4a_1 a_2(c_2-c_1)+
a_2 b_1^2-a_1 b_2 ^2=0\;. 
\] 
Show that, in the case  $a_1 \ne a_2 \,$, the two curves have exactly one common tangent if and only if they touch each other. In the case $a_1 =a_2\,$, find a necessary and sufficient condition for the two curves to have exactly one common tangent.

Solution source

\begin{align*}
&& y' &= 2ax+b \\
\Rightarrow && m &= 2ax_t+b \\
\Rightarrow && x_t &= \frac{m-b}{2a}
\end{align*}

Therefore we must have 
\begin{align*}
mx_t &= 2ax_t^2+bx_t \\
y - mx &= ax_t^2+bx_t+c - mx_t \\
&= ax_t^2+bx_t+c - (2ax_t^2+bx_t) \\
&= c - ax_t^2 \\
&= c-a\left (\frac{m-b}{2a} \right)^2 \\
&= c - \frac{(m-b)^2}{4a}
\end{align*}

They will have a common tangent if and only if the constant terms are equal, ie

\begin{align*}
&& c_1 - \frac{(m-b_1)^2}{4a_1} &= c_2 - \frac{(m-b_2)^2}{4a_2} \\
\Leftrightarrow && (c_1-c_2) &= \frac{(m-b_1)^2}{4a_1} -\frac{(m-b_2)^2}{4a_2} \\
\Leftrightarrow && 4a_1a_2(c_1-c_2) &= a_2(m-b_1)^2-a_1(m-b_2)^2 \\
&&&= (a_2-a_1)m^2+2(a_1b_2-a_2b_1)m+a_2b_1^2-a_1b_2^2
\end{align*}

as required.

Treating this as a polynomial in $m$, we can see that the two curves will have exactly one common tangent iff $\Delta = 0$, ie:

\begin{align*}
&& 0 &= \Delta \\
&&&= (2(a_1b_2-a_2b_1))^2 - 4 (a_2-a_1)(4a_1 a_2(c_2-c_1)+
a_2 b_1^2-a_1 b_2 ^2) \\
&&&= 4a_1^2b_2^2-8a_1a_2b_1b_2+4a_2b_1^2 - 4a_2^2b_1^2-4a_1^2b_2^2 + 4a_1a_2(b_1^2+b_2^2)-16(a_2-a_1)a_1a_2(c_2-c_1) \\
&&&=-8a_1a_2b_1b_2+4a_1a_2(b_1^2+b_2^2)-16(a_2-a_1)a_1a_2(c_2-c_1) \\
&&&=a_1a_2(4(b_1-b_2)^2-16(a_2-a_1)(c_2-c_1)) \\
&&&= 4a_1a_2((b_2-b_1)^2 - 4(a_2-a_1)(c_2-c_1)
\end{align*}

But this is just the discriminant of the difference, ie equivalent to the two parabolas just touching. (Assuming $a_1-a_2 \neq 0$ and we do end up with a quadratic).

If $a_1 = a_2 = a$ then we need exactly one solution to $2a(b_1-b_2)m +4a^2(c_2-c_1)+a(b_1^2-b_2^2) = 0$, ie $b_1 \neq b_2$.