Let the point of intersection be P(h,k)
Equation of tangent to y2=4ax in slope form is
y=mx+ammy=m2x+am2x−my+a=0
It passes through P(h,k)
m2h−mk+a=0
The equation is quadractic in m
∴m1+m2=−−kh=kh.....(i)m1m2=ah.........(ii)
Slope of line which makes equal angle with the tangents =tanα
m1−tanα1+m1tanα=tanα−m21+m2tanαm1+m1m2tanα−tanα−m2tan2α=tanα−m2+m1tan2α−m1m2tanαm1+m2−(m1+m2)tan2α−2tanα+2m1m2tanα=0(1−tan2α)(m1+m2)−2tanα+2tanαm1m2=0(m1+m2)−2tanα1−tan2α+2tanα1−tan2αm1m2=0(m1+m2)−tan2α+tan2αm1m2=0
using (i) and (ii)
kh−tan2α+tan2αah=0k−htan2α+atan2α=0k=htan2α−atan2α
Replacing h by x and k by y
y=tan2α(x−a)