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 $f(x)=\frac{1}{8}x^{2}-2$.
