Tangents to a parabola that go through the same point

Question

The question is: The two lines tangent to f (x) = $x^2$ + 4x + 2 through the point (2, -12)have equations y = ax + b and y = cx + d, respectively. What is the value of a + b + c + d?

What I did to solve it: f '(x) = 2x + 4. The point is (2 -12) so I plugged in two to get f '(2) = 8. I used this in point-slope form to get y + 12 = 8(x-2) => y = 8x -28. Since it is two equations, I added them up and multiplied by 2 and got -40, which is the answer. I feel like this is a fluke and it doesn't make sense to me why this worked. Can someone explain either why it works or a way that will always work? Thanks in advance!

Answer 1

This is no coincidence, or if it is, it’s a happy one.  

Let $P_0=(x_0,y_0)$ be the given external point and $x_1$ and $x_2$ the ordinates of the points of tangency. The equations of the two tangent lines are $y=f'(x_1)(x-x_0)+y_0$ and $y=f'(x_2)(x-x_0)+y_0$. On the other hand, the equation of the line that you found is $y=f'(x_0)(x-x_0)+y_0$. So, for the given parabola, we have $$f'(x_1)+y_0-f'(x_1)x_0+f'(x_2)+y_0-f'(x_2)x_0 = 2(f'(x_0)+y_0-f'(x_0)x_0)$$ or $$(1-x_0)(f'(x_1)+f'(x_2))+2y_0 = 2(1-x_0)f'(x_0)+2y_0$$ which reduces to $$f'(x_1)+f'(x_2)=2f'(x_0).\tag{1}$$  

As it turns out, this holds for any parabola given by the general formula $f(x)=ax^2+bx+c$. The ordinates of the points of tangency are the solutions to $$f(x)-y_0=f'(x)(x-x_0),$$ which after expanding, rearranging and simplifying becomes $$x^2-2x_0x+{y_0-bx_0-c\over a}=0.\tag{2}$$ On the other hand, substituting the general formula into (1) yields $x_1+x_2=2x_0$, which can be verified by a glance at (2).  

Observe that this last equation tells us that $x_0$ is midway between $x_1$ and $x_2$, which means that the chord connecting the two tangent points is parallel to the tangent at $x=x_0$. This follows from general properties of parabolas, but can be verified directly: $$\begin{align} f(x_2)-f(x_1) &= f'(x_2)(x_2-x_0)-f'(x_1)(x_1-x_0) \\ &= (2ax_2+b)(x_2-x_0)-(2ax_1+b)(x_1-x_0) \\ &= (2a(x_1+x_2-x_0)+b)(x_2-x_1) \\ &= (2ax_0+b)(x_2-x_1) \end{align}$$ therefore $${f(x_2)-f(x_1)\over x_2-x_1}=2ax_0+b=f'(x_0).$$ A similar derivation demonstrates that for an arbitrary chord, the tangents at its endpoints intersect at a point with ordinate halfway between the ordinates of those endpoints. Combining this with (1), we find that the slope of a chord is equal to the average of the slopes of the tangents at its endpoints.

Answer 2

Look at the figure:

The point $A=(2,-12)$ is not a point of the parabola, so the slopes of the tangents to the parabola from this point cannot be be simply derived starting from the derivative of the parabola at $x=2$.

If $P=(X,X^2+4X+2)$ is a point of tangency (the points $C$ and $D$ in the figure), than the slope $m=(y_P-y_A)/(x_P-x_A)$ of the line $PA$ must be the same as the slope of the tangent to the parabola in $P$, i.e.: $m=f'(x_P)$ . So, For our $A$ and $P$ we have:

$$ \frac{X^2+4X+2+12}{X-2}=2X+4 $$

Solving this equation we find the coordinates $x_C$, $x_D$ of the two points of tangency, from wich we can find the equations of the two tangent lines.

Another simple solution, without using derivative, is to note that the system: $$ \begin {cases} y=x^2+4x+2\\ y+12=m(x-2) \end{cases} $$

represents the intersection between the parabola and the lines that passe thorough $A$, and a line is tangent to the parabola if the system has only one (double) solution. This means that the discriminat $\Delta(m)$ of the second degree equation that solve the system is such that: $\Delta(m)=0$ .

Since $\Delta(m)$ is a second degree polynomial in the parameter $m$, this is a second degree equation in $m$ that gives the two values of $m$ for the slopes for the two tangent lines.