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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,2012-04-15

1 Answers 1

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

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.2012-04-15