It’s easy to find the equation of the chord using $$T=S_1$$ $$2ax-ky +2ah-4h+k^2=0$$ where (h,k) are midpoints of the chord
I identified to possible ways to solve this. Either solve this equation and the equation of the parabola, obtain the points of intersection in terms of h and k, find the distance between the two points and equate it to 2l
Or
Use the coordinates $(t_1)$ and $(t_2)$ on the parabola, which would given the midpoints, and then finding the distance.....
I tried both of them, but they are so awfully long, I am convinced there is proper, much shorter method to solve this. How should I solve it correctly?
The answer is $(4x-y^2)(y^2+4)=4l^2$
Note that $a =1$. So, the chord equation reduces to,
$$2x-ky +k^2-2h=0$$
The two intersection points with the parabola can then be found by substituting $y^2=4x$ into the equation,
$$y^2 -2ky+2k^2-4h=0$$
which leads to
$$y_1+y_2 = 2k, \>\>\>\>\>y_1y_2 = 2k^2-4h$$
Evaluate,
$$(y_2-y_1)^2 = (y_2+y_1)^2 -4y_1y_2 = 4(4h-k^2)$$ $$(x_2-x_1)^2 = (\frac1{4}y_2^2-\frac1{4}y_1^2)^2= \frac1{16}(y_1+y_2)^2(y_1-y_2)^2 = k^2(4h-k^2)$$
Then, matching the length of the chord $(x_2-x_1)^2 +(y_2-y_1)^2 = 4l^2$ to get the locus,
$$ k^2(4h-k^2) + 4(4h-k^2) = 4l^2$$
or, in the form,
$$(4h-k^2)(k^2+4)=4l^2$$