pick point on parabola so 2 conditions are true

Question

You have a parabola

$$y=ax^2+bx+c $$

We know $a,b$ & $c$. On this parabola you have to pick a point A where the following conditions are true:

1) If you draw a tangent line in this point to the parabola, then the middle point of the following 3 points has to be the midpoint of the other two points (same distance):

• the point where the parabola touches the tangent line (point $A$)

• the intersection of the tangent line with the horizontal line drawn through the vertex: $$y=\frac{4ac-b^2}{4a}.$$

• the intersection of the tangent line with the axis of symmetry = vertical line through the vertex: $x=\dfrac{-b}{2a}$.

2) Normally, then also for the following $3$ points the middle point is on the same distance to the first and to the third point (all be it on a different distance as Condition 1, depending on the shape of the parabola).

• the intersection point of the horizontal line through point A and the axis of symmetry

• the vertex

• the intersection point of the tangent line to the parabola in point $A$, with the axis of symmetry

I know that the slope of the tangent line in a certain $Ax$ will be $2a Ax+b$, so you can calculate the tangent line through $A$ to be equal to $y=(2 a A x+ b)(x-Ax)+Ay $, but I'm not sure how to move from there to pick point $A$ so all conditions are still true.

All help is truly appreciated. These things can be pretty frustrating if you're not mathematically inclined.

In this image, pick point A such that $|AF| = |FE|$ and $|ED| =|DB|$. I can do it geometrically, but apparently not algebraically.

Answer 1

Let $A=(x_1,y_1),$ so that $y_1=ax_1^2+bx_1+c.$ Observe that if $x_1=\frac{-b}{2a},$ then the three points coincide, and we're done, so suppose not. Then $2ax_1\ne -b,$ so $2ax_1+b\ne0.$

Your tangent line will be $$\begin{align}y &= (2ax_1+b)(x-x_1)+y_1\\ &= (2ax_1+b)(x-x_1)+ax_1^2+bx_1+c\\ &= (2ax_1+b)x-2ax_1^2-bx_1+ax_1^2+bx_1+c\\ &= (2ax_1+b)x-ax_1^2+c\end{align}$$ then. This intersects the axis of symmetry at point $E=(x_3,y_3)$ where $x_3=\frac{-b}{2a},$ so that $$y_3=(2ax_1+b)x_3-ax_1^2+c=(2ax_1+b)\cdot\frac{-b}{2a}-ax_1^2+c=-ax_1^2-bx_1+c+\frac{b^2}{2a}.$$

Your tangent line will intersect the horizontal line through the vertex at the point $F=(x_2,y_2),$ where $y_2=c-\frac{b^2}{4a}.$ Then $$c-\frac{b^2}{4a}=(2ax_1+b)x_2-ax_1^2+c\\ax_1^2-\frac{b^2}{4a}=(2ax_1+b)x_2\\\frac{4a^2x_1^2-b^2}{4a}=(2ax_1+b)x_2\\\frac{(2ax_1+b)(2ax_1-b)}{4a}=(2ax_1+b)x_2\\\frac{2ax_1-b}{4a}=x_2.$$

In this way, we've determined all three points in terms of $a,b,c,x_1.$ Your task is to determine $x_1$ so that $F$ is the mid point of $A$ and $E$--that is, $x_2=\frac{x_1+x_3}2$ and $y_2=\frac{y_1+y_3}2.$ You should see that this is always the case--regardless of our choice of the point $A$!

A similar approach can be taken for the other problem.

Answer 2

Analytical solution outline

Through rotations and translations, you can bring any parabola to the origin, solve this problem, and then rotate and translate back to the original position to find the solution point. So, it suffices that we solve the problem for $y=ax^2$. Suppose the point of tangency is $(p,q)$. Under the circumstances, both $a,p\neq 0$.

Then the tangent line is given in point-slope form as $y-q=2ap(x-p)$. The other two points lie on this line, given by substituting $x=0$ and $y=0$.

Thus the three points are, in order: $(p,q)$, $(0,q-2ap^2)$, $(\frac{2ap^2-q}{2ap},0)$.

With a lot of patience, or else the aid of an algebra program, you can show the distance between the first two to be $\sqrt{(q-2ap^2)^2+(p-\frac{apq}{2})^2}$ and the distance between the second two to be $\sqrt{\frac{q^2(a^2p^2+4)}{4}}$.

If these two distances are to be equal, the resulting equation is $q=\frac{4a^2p^2+1}{5a}$. Finally, $(p,q)$ must lie on the parabola: that is, $q=ap^2$.

Solving $ap^2=\frac{4a^2p^2+1}{5a}$, you get that $p=\frac{1}{a^2}$, and consequently $q=\frac{1}{a^3}$

To get this to be a general solution, you would just rotate and translate the parabola into position, rotating and translating this point along with it.