# 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!

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

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+1: nice graph... – Fabian Apr 6 '11 at 21:20
What software did you use for the graph? – Bill Dubuque Apr 7 '11 at 1:12
@Bill, It is SWP (Scientific Work Place). – Américo Tavares Apr 7 '11 at 9:14
Perfect, thankyou! – MathsStudent Apr 7 '11 at 11:35

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 $a, b$.

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