Certain chords of a parameterized parabola pass through a common point

Question

If the end points $P(t_1)$ and $Q (t_2)$ of a chord of a parabola with parameterization $(at^2, 2at)$ satisfy the relation $t_1t_2 = k$ (constant) then we have to find the fixed point through which chord will always pass.

I tried

I wrote the equation of chord i.e $2x - (t_1+ t_2) + 2at_1t_2 = 0$

After putting $t_2= \frac{t_1}{k}$

I got $2x -t_1y - k(\frac{y}{t_1} -2a) = 0$

But how do I will get the fixed point.

Answer 1

The chord joining the points $P'(-t_1), Q'(-t_2)$ must also pass through this fixed point, so by symmetry the fixed point has $y=0$ and therefore the coordinates are $$(-ka, 0)$$

Note that in the special case where $k=-1$ the fixed point is the focus $(a, 0)$