The correct option is
A y2=a(x−3a)The given parabola is
y2=4axLet t denote the parameter, then the equation of the normal to this parabola is given by y=−tx+2at+at3
Now this normal passes through the point (h,k) then
k=−th+2at+at3⇒at3+t(2a−h)−k=0 ....... (A)
Let m1,m2,m3 be the roots of this equation.
Let the perpendicular normals correspond to the values of m1,m2
So, m1m2=−1
Now from equation (A), m1m2m3=ka
m1m2=−1 gives m3=−ka
Since m3 is a root of equation (A), we have
a(−ka)3−ka(2a−h)−k=0⇒k2=a(h−3a)
So locus of (h,k) is y2=a(x−3a)