Normals are drawn to the parabola ${\text{y}}^2=4x$  from any point on the line y = 2, then vertices of the triangle formed by tangents at feet of perpendiculars lie on a fixed rectangular hyperbola  $xy=-c^2$ where $c^2$ is ________

Equation of normal at  $(t^2,2t)$ is  $y=-xt+2t+t^3$
Let any point on  $y=2$  is  $(h,2)$
If normal passes through (h,2)
 $2+ht-2t-t^3=0$
 $t^3-t(\mathrm{h}-2)-2=0$
$t_1,t_{2}and{t}_3$ are points of this equation vertices lies on  $xy=-2$
This is because tangents at  $t_{1}and{t}_2$ meet a  $(t_1,t_2,t_1+t_2)=(2/t_3,-t_3)$