Finding coefficients in polynomial given three tangents

Question

I am stuck with a problem I simply cannot solve.

I have to find the coefficients of a quadratic polynomial given three tangents. The problem is stated as follows:

The three lines described by the equations

$y_1(x)=-4x-16.5$

$y_2(x)=2x-4.5$

$y_3(x)=6x-16.5$

are all tangents to a quadratic polynomial $p(x)=ax^2+bx+c$

Determine the values of the coefficients a, b & c.

I simply cannot solve this problem, I've been at it for a long time. Any help is greatly appreciated :)

Edit: I'm including the way I tried to solve it. I didn't get super far.

Given the polynomial $p(x)$ I know that $p'(x)=2ax+b$

Therefore, the following is true for the three points with x-values of $x_1, x_2 $ and $x_3$, where the lines $y_1, y_2$ and $y_3$ are tangent to the parabola:

$-4=2ax_1+b$

$2=2ax_2+b$

$6=2ax_3+b$

That's all I've managed to do. I've also found the points where the three lines intersect (well, three points two of the lines intersect), but I can't think of how to use that for anything.

Answer 1

Note that if a line and a quadratic are tangent $mx+d=ax^2+bx+c$ then the following quadratic will have discriminant zero \begin{eqnarray*} ax^2+(b-m)x+c-d=0. \end{eqnarray*} This will lead to $3$ equations for $a,b,c$ that are easily solved giving

$(a,b,c)=(1/2,1,-4)$.

Answer 2

If we have a parabola $$ y = a x^2 + b x + c $$ and a line $$ y = mx + d,$$ they are tangent if $$ b^2 - 4ac = -4da + 2mb - m^2 $$ as then the parabola $$ y = a x^2 + (b-m) x + (c -d) $$ has a double root, i.e. is a constant times a square, $$ a (x+p)^2 $$ Might as well write this: I fixed $$ \Delta = b^2 - 4 a c $$ and then had three equations $$ -4da + 2mb = m^2 + \Delta $$ by plugging in the values from the three lines $ y = mx + d.$ I expected bigger problems, but just taking the differences of two of the equations cancels the extra unknown $\Delta,$ alowing us to find $a,b$ quickly. From that, we finally get a value for $\Delta,$ after which we have one equation for $c$

$$ 66a - 8 b = 16 + \Delta \; , $$ $$ 66a + 12 b = 36 + \Delta \; , $$ $$ 18a + 4 b = 4 + \Delta \; . $$ Second minus first gives $b.$ Plug in the $b$ value, then subtract second minus third, which gives $a.$ Plug both into any of the three to find $\Delta.$ Finally, $c = \frac{b^2 - \Delta}{4a}$

Answer 3

Given any two tangents to a parabola of the form $y = f(x) = ax^2 + bx + c,$ the $x$-coordinate of the intersection of the tangent lines will be midway between the $x$-coordinates of the tangent points.

Working out the intersection points of the three lines, we can see that the intersection of $y = y_1(x)$ and $y = y_2(x)$ occurs at $x = -2$ and the intersection of $y = y_2(x)$ and $y = y_3(x)$ occurs at $x = 3.$ Let

• $x = x_1$ at the tangent point with $y = y_1(x)$;
• $x = x_2$ the tangent point with $y = y_2(x)$; and
• $x = x_3$ the tangent point with $y = y_3(x).$

Due to the $y$-coordinates and slopes at the intersection points, it is clear that $x_1 < -2 < x_2 < 3 < x_3$; moreover, $-2$ is midway between $x_1$ and $x_2$ and $3$ is midway between $x_2$ and $x_3.$ It follows that $$x_3 - x_1 = 2\left(\frac{x_3 + x_2}2 - \frac{x_2 + x_1}2\right) = 2(3 - (-2)) = 10.$$

The tangency condition implies that $f'(x_1) = y_1'(x) = -4$ and $f'(x_3) = y_3'(x) = 6.$ But $f'(x) = 2ax + b,$ so $20a = 2a(x_3 - x_1) = f'(x_3) - f'(x_1) = 10,$ and therefore $a = \frac12.$ It is then a relatively straightforward exercise to find the other two coefficients.