Since the two pairs of tangents are symmetric with respect to the $y$-axe,
the quadratic function $f(x)=ax^{2}+bx+c$ must be even ($f(x)=f(-x)$),
which implies that $b=0$. The equations of the tangents to the graph of $f(x)=ax^{2}+c$ at points $%
\left( x_{1},f(x_{1})\right) $ and $\left( x_{2},f(x_{2})\right) $ are
$$\begin{eqnarray*}
y &=&f^{\prime }(x_{i})x-f^{\prime }(x_{i})x_{i}+f(x_{i})\qquad i=1,2 \\
&=&2ax_{i}x+c-ax_{i}^{2}.
\end{eqnarray*}$$
These equations must be equivalent to two of the given tangents, one from each pair, e.g. $y=2x-10$ and $y=x-4$:
$$\left\{
\begin{array}{c}
2ax_{1}x+c-ax_{1}^{2}=2x-10 \\
2ax_{2}x+c-ax_{2}^{2}=x-4%
\end{array}%
\right. $$
Finally we compare coefficients and solve the resulting system of $4$
equations:
$$\left\{
\begin{array}{c}
2ax_{1}=2 \\
c-ax_{1}^{2}=-10 \\
2ax_{2}=1 \\
c-ax_{2}^{2}=-4%
\end{array}%
\right. \Leftrightarrow \left\{
\begin{array}{c}
x_{1}=8 \\
x_{2}=4 \\
a=\frac{1}{8} \\
c=-2%
\end{array}%
\right. $$
Thus the quadratic is $f(x)=\frac{1}{8}x^{2}-2$.
