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OK, I wasn't on a class regarding this type of excercises. I got the notes from the lesson but have no idea how is it working. I hope you'll be able to clarify:

Determine the equation of the curve being a set of the middles of all the chords of the parabola $y=x^2-2$ going through (0,0). Sketch this curve.

let $y=mx$ the line through the origin, then we get $$x^2-mx-2=0$$ Solving this equation we obtain $$x=\frac{m}{2}+\sqrt{\frac{m^2}{4}+2}$$ $$y=m(\frac{m}{2}+\sqrt{\frac{m^2}{4}+2})$$ and $$x=\frac{m}{2}-\sqrt{\frac{m^2}{4}+2}$$ $$y=m(\frac{m}{2}-\sqrt{\frac{m^2}{4}+2})$$ thus the middle point of the chords is given by $M(m;m^2)$ and this equation is $y=x^2$.

..and all seems to be quite obvious but I have not the slightest idea why the middle moint is given by $M(m;m^2)$. How do the values of x'es an y'es we obtained imply that the middles of x and y are such and that the equation is y=$x^2$? Could you please help?

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middle point should be $M(\frac{m}{2},\frac{m^2}{2})$ and equation $y=2x^2$ –  pedja Oct 23 '11 at 17:56
    
..which doesn't change the fact I don't know what's going on. Could you please help? –  Bringiton Oct 23 '11 at 17:57

2 Answers 2

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You procedure is correct. Let $y=mx$ be a line through the origin. Then the $x$-coordinates of the points at which this line meets the parabola are the solutions of the equation $x^2-mx-2=0$.

Recall that the midpoint of the line segment that joins $(a,b)$ and $(c,d)$ has coordinates $$\left(\frac{a+c}{2}, \frac{b+d}{2}\right).$$ (The $x$-coordinate is the average of the two $x$-coordinates, and the $y$-coordinate is the average of the two $y$-coordinates.)

So to find the $x$-coordinate of the midpoint, we can find the $x$-coordinates of the points of intersection, and then find their average. The work you did shows that you are well on your way to doing that.

However, there is a shortcut. Recall, perhaps, that the sum of the solutions of the quadratic equation $ax^2+bx+c=0$ is $-b/a$. So the sum of the solutions of $x^2-mx-2=0$ is $m$.

If you do not recall this result, let's derive it. Note that if $r$ and $s$ are the roots of a quadratic polynomial $P(x)$, and the coefficient of $x^2$ is $1$, then $P(x)=(x-r)(x-s)=x^2-(r+s)x+rs$. So the sum of the roots is the negative of the coefficient of $x$, and the product of the roots is the constant term. This result, and generalizations to polynomials of higher degree, is very useful.

Since the sum of the $x$-coordinates of our points of intersection is $m$, the average of this sum is $\frac{m}{2}$. We did not even have to solve the quadratic equation to find the answer!

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I think I get it now. Thank you a lot! :) –  Bringiton Oct 23 '11 at 18:24

In order to find middle point you have to add those two expressions for $x$ and for $y$:

$$x_M=\frac{\frac{m}{2}+\sqrt{\frac{m^2}{4}+2}+\frac{m}{2}-\sqrt{\frac{m^2}{4}+2}}{2}=\frac{m}{2}$$

$$y_M=\frac{m(\frac{m}{2}+\sqrt{\frac{m^2}{4}+2})+m(\frac{m}{2}-\sqrt{\frac{m^2}{4}+2})}{2}=\frac{m^2}{2}$$

Now,we may write:

$y=2\frac{m^2}{4}=2(\frac{m}{2})^2=2x^2$ ,so $y=2x^2$ is equation of the curve that contains all middle points.

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I think I get it now. Thank you :) –  Bringiton Oct 23 '11 at 18:24

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