Find the given locus

Let the points be $(at_{1}^{2},2at_{1})$ and $(at_{2}^{2},2at_{2})$

Equation of chord joining these points is

$\large{y-2at_{1}=\frac{2}{t_{1}+t_{2}}(x-at_{1}^{2})}$

it passes through $(3a,a)$

$\large{a-2at_{1}=\frac{2}{t_{1}+t_{2}}(3a-at_{1}^{2})}$

$t_{1}+t_{2}-2t_{1}t_{2}=6\quad \quad \quad (1)$

Let $(h,k)$ be the mid-point

$\large{\frac{at_{1}^{2}+at_{2}^{2}}{2}=h}$

$\large{\frac{2at_{1}+2at_{2}}{2}=k}$

$\large{t_{1}^{2}+t_{2}^{2}=2\frac{h}{a}\quad \quad \quad (2)}$

$\large{t_{1}+t_{2}=\frac{k}{a}\quad \quad \quad (3)}$

Also $(t_{1}+t_{2})^{2}-2t_{1}t_{2}=2\frac{h}{a}$

From $(1),(2),(3)$

$\large{\frac{k^{2}}{a^{2}}-(\frac{k}{a}-6)=2\frac{h}{a}}$

$k^{2}-2ah-ak+6a^{2}=0$

Use the fact that equation of chord whose midpoint is say (x,y) is given by the equation

T = S1 .

Now pass this from (a,3a) to get required locus . As simple as that :)