If the locus of centres of a family of circles passing through the vertex of the parabola $y^2=4ax$ and cutting the parabola orthogonally at the other point of intersection is $2y^2\left(2y^2+x^2-12ax\right)=ax{\left(kx-4a\right)}^2$ , then the value of $k$ is 

Let P (at² , 2at) be any point on y²= 4ax. Then vertex A(0, 0). The equation of tangent at P is ty = x +at² …. (1) Tangent at P to the parabola will be normal to the circle; AP is a chord whose mid point is $\left(\frac{{\mathrm{at}}^2}{2},\mathrm{at}\right)$ and slope is 2/t. Equation of the line passing through midpoint of AP and perpendicular to AP is y - at = $\frac{-\mathrm{t}}{2}\left(\mathrm{x}-\frac{{\mathrm{at}}^2}{2}\right);$
$\mathrm{tx}+2\mathrm{y}=\frac{{\mathrm{at}}^3}{2}+2\mathrm{at}\left(2\right)$
(1) And (2) both pass through (x₁, y₁) which is the centre of the circle 
ty₁ = x₁ + at₂…..(3) 2tx₁ + 4y₁ = at³ + 4at….(4) 
we have T² y₁ + t(4a-3x₁)-4y₁ = 0…(5) 
Also from (3), at² -ty₁+x₁=0….(6) 
Eliminating t from (5) &(6)
$\begin{array}{l}\frac{{\mathrm{t}}^2}{{\mathrm{x}}_1\left(4\mathrm{a}-3{\mathrm{x}}_1\right)}=\frac{\mathrm{t}}{-4{\mathrm{ay}}_1-{\mathrm{x}}_1{\mathrm{y}}_1}\\ =\frac{1}{-{\mathrm{y}}_1^2-\mathrm{a}\left(4\mathrm{a}-3{\mathrm{x}}_1\right)}\end{array}$
On simplyfing we get
$2{\mathrm{y}}_1^2\left(2{\mathrm{y}}_1^2+{\mathrm{x}}_1^2-12\mathrm{ax}\right)=\mathrm{ax}(3\times -4\mathrm{a}{)}^2$
Hence required locus is