Points $A$ and $B$ lie on the parabola $y=2x^2+4x-2,$such that the origin is the mid point of the line segment $AB$

Question

Points $A$ and $B$ lie on the parabola $y=2x^2+4x-2,$such that the origin is the mid point of the line segment $AB$.Find the length of the line segment $AB$

---

$y=2(x^2+2x-1)=2(x+1)^2-4\implies (y+4)=2(x+2)^2$ and let $x=t-2,y=2t^2-4$ be the parametric equation of the parabola.
I am stuck here.

Answer 1

Let the points be $A(x_1,2x_1^2+4x_1-2)$ and $B(x_2,2x_2^2+4x_2-2)$. Then: $$\begin{cases}\frac12(x_1+x_2)=0\\ \frac12(2x_1^2+4x_1-2+2x_2^2+4x_2-2)=0\end{cases} \Rightarrow (x_1,x_2)=(\pm 1,\mp 1).$$ Hence: $A(1,4)$, $B(-1,-4)$ and $AB=\sqrt{(-1-1)^2+(-4-4)^2}=\sqrt{68}$.

Answer 2

What does the condition, that the mid point of $AB$ is the origin, say about the two points $A$ and $B$? How can you express one by the other? Then you can solve the equation for two unique points.

Answer 3

I’m pretty sure that there must be a more clever approach than the following, but one way to attack this problem is to use the fact that the midpoints of parallel chords all lie on a line parallel to the parabola’s axis. Therefore, the slope of $AB$ is equal to the slope of the tangent to the parabola at $x=0$. That gets you the equation of this line, from which you can find the coordinates of $A$ and $B$.