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The question was:

The points P and Q on the curve: $$x = 2at, y= at^2$$ have parameters p and q respectively. Show that PQ intersects the directrix at: $$ \left (\frac{2a(pq-1)}{p+q},-a \right ) $$

I've managed to find that the equation of the chord PQ is: $$ y - \frac{1}{2} (p+q)x+apq=0 $$ but after this I'm a bit confused has to how to find the directrix using a parametric equation.

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Welcome to MSE. +1 for showing us what you know. Please continue to do the same. Regards, – user21436 Apr 15 '12 at 13:42
up vote 0 down vote accepted

You're there. To proceed, here's a hint.


This is a standard parabola given by $x^2=4ay$ and its directrix is at $y=-a$. So,

to solve for the point of intersection $PQ$ with $y=-a$ is to just plug in $y=-a$ and solve the resulting linear equation in $x$ (for $x$, of course.)

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Thank you very much for your answer! I now understand how to find the equation of the directrix. – Jallah Apr 15 '12 at 13:45

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