Prove that the chord always passes through a fixed point

Question

If the end points $P(t_1)$ and $Q(t_2)$ of a chord of a parabola $y^2=4ax$ satisfy the relation $t_1t_2=k$(constant) then prove that the chord always passes through a fixed point.Find that point also.

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Let $P(at^2_1,2at_1)$ and $Q(at^2_2,2at_2)$,then the equation of the line joining P and Q is $y(t_1+t_2)-2x-2at_2=0$
But i am stuck here.Please help.

Answer 1

Let the parameters be $p,q$ (instead of $t_1, t_2$) hence points are $P(ap^2, 2ap), Q(aq^2, 2aq)$.

Equation of chord PQ:

$$\begin{align} \frac {y-2ap}{2aq-2ap}&=\frac {x-ap^2}{aq^2-ap^2}\\ \frac {y-2ap}{2(q-p)}&=\frac {x-ap^2}{(q-p)(q+p)} &&(q\neq p)\\ (p+q)y&=2(x+apq)\\ (p+q)y&=2(x+ka)\end{align}$$ which always passes through $(-ka, 0)$.

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It is interesting to note the following:

• when $k=-1$, the chord always passes through the focus, $(a,0)$,

• when $k=1$, the chord always intersects the $x-$axis at the directrix, i.e. at $(-a,0)$.