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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.

user984325
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3 Answers3

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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 $B$.

enter image description here

amd
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  • ,If $A(u-2,2u^2-4),B(v-2,2v^2-4)$,then slope of $AB=2(u+v)$ and the slope of the tangent to the parabola at $x=0$ is $4$,equating these gives $u+v=2$ but origin is also the mid point of $AB$ which gives $u+v=4$,it is not possible. – user984325 Nov 27 '18 at 02:02
  • @user984325 And yet, the attached illustration shows a perfectly good solution to your problem. The origin is clearly the midpoint of $AB$. Your parameterization is incorrect. Try again. – amd Nov 27 '18 at 03:44
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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 $AB=\sqrt{(-1-1)^2+(-4-4)^2}=\sqrt{68}$.

farruhota
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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.

Monoidoid
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