The correct option is A y2=a(x−a)
Eqaution of normal to y2=4ax passing through P whose locus is (h,k) is
k=mh−2am−am3
⇒am3+(2a−h)m+k=0 ⋯(1)
It has 3 roots.
m1m2m3=−ka
As 2 tangents make complementary angles,
so if m1=tanθ, then m2=cotθ
∴ m3=−ka
Putting this in equation (1),
a(−ka)3+(2a−h)(−ka)+k=0
By solving, we get
k2=ah−a2
∴y2=a(x−a)