Find equation of quadratic when given tangents? I know the equations of 4 lines which are tangents to a quadratic:
$y=2x-10$
$y=x-4$
$y=-x-4$
$y=-2x-10$
If I know that all of these equations are tangents, how do I find the equation of the quadratic?
Normally I would be told where the tangents touch the curve, but that info isn't given.
Thanks!
 A: 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 $f(x)=\frac{1}{8}x^{2}-2$.

A: As they are symmetric around the origin, the quadratic has no linear term in $x$.  So I would put $y^2=ax^2+b$ as any linear term in $y$ can be absorbed into a vertical shift or $y=ax^2+b$ to get the parabolas.  Then calculate what $a$ and $b$ need to be to make them tangent.  Because we incorporated the symmetry, you only have two tangent lines, but that gives two equations for $a, b$.
