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If $PQ$ is a focal chord of the parabola $x^2=4py$ and the coordinates of $P$ are $(x_{0}, y_{0})$ show that the coordinates of $Q$ are $$\left(\frac{4p^2}{x_0}, \frac{p^2}{y_0}\right)$$

I labeled the focus of the parabola $O$. I tried to find the coordinates of $Q$ by using the slope of $PO$, which should be equivalent to the slope of $OQ$. I got $(-x_0, -y_0 + 2p)$ as a result, which doesn't seem close to what I have to show. I also tried plugging in $\frac{x_0^2}{4p}$ into $y_0$ (just solved for y in the original equation of the parabola), but that didn't get me too far either. Can I use this approach at all?

Also, quick additional question, whenever I search up things about parabolas I see parametric equations a lot, could they be used here?

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HINT:

Any point of the parabola can be written as $\displaystyle Q(2pt,pt^2)$

Also, the focus is $\displaystyle O(0,a)$

One endpoint being $\displaystyle P(x_0,y_0)$

As $P,O,Q$ are co-linear $\triangle POQ=0 $.

Use this to find $t$


Alternatively, the gradient of $OQ=$ that of $PQ$

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  • $\begingroup$ Thanks, I've seen $Q(2pt, pt^2)$ show up a few times, not in my textbook though (precalc, but I'll have another look). Where did it come from? $\endgroup$ – Howcan Feb 25 '14 at 10:04
  • $\begingroup$ @Howcan, $$x^2=4py=\iff \left(\frac x{2p}\right)^2=\frac yp, $$ Set $$\frac x{2p}=t, \frac yp=?$$ $\endgroup$ – lab bhattacharjee Feb 25 '14 at 10:06
  • $\begingroup$ I understood that, I'm still not sure how to use them. I tried to show that the three points would make a triangle with an area of 0 and solve for t. Is that what I should do? If yes then I just made some algebraic error. $\endgroup$ – Howcan Feb 25 '14 at 10:23
  • $\begingroup$ Also, what do you mean by gradient? Do you just mean the slope? If so I tried doing that in my question and it didn't seem to work. Unless you meant gradient of vectors which I know nothing about. $\endgroup$ – Howcan Feb 25 '14 at 10:47
  • $\begingroup$ @Howcan, mathsisfun.com/gradient.html $\endgroup$ – lab bhattacharjee Feb 25 '14 at 10:56

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