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I need help with this question that i attempted to solve using the equation $y^2=4ax$ :

"Prove that on the axis of any parabola there is a certain point which has the property that,if a chord PQ of the parabola be drawn through it,then :-

$1/PK^2/$ + $1/QK^2$ is the same for all positions of the chord."

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I use the parametric equations of the parabola.

Then let $P\equiv(at^2,2at)$, $Q\equiv(a{t_1}^2,2at_1)$ and $K\equiv(c,0)$.

Because $P,Q$ and $K$ are collinear, one has $$\frac{2at-2at_1}{at^2-a{t_1}^2}=\frac{2at}{at^2-c}$$ hence $$t_1=-\frac c{at}$$ Then $$\frac 1{PK^2}+\frac 1{QK^2}=\frac {c^2+a^2t^4}{c^2\,[c^2+(4a^2-2ac)\,t^2+a^2t^4]}$$

Putting $\,c=2a\,$, one obtains $$\frac 1{PK^2}+\frac 1{QK^2}=\frac 1{4a^2}$$

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Hint

Note that if $K$ is at $(X,0)$ and we choose the $x$-axis as the chord we get $\frac 1{X^2}$ for the sum.

Now choose the line $x=X$ as the chord, the sum is also easy to calculate.

This should enable you to identify the point $K$. From there it should be relatively easy computation.

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