The correct option is A y2=a(x−3a)
Let P(x1,y1) be the point of intersectin of the two perpendicular normals at A(t1),B(t2) on the parabola y2=4ax
let t3 be the foot of the third normal through P.Equation of a normal at t to the parabola is y+xt=2at+at3
If this normal passes through P then y1+x1t=2at+at3⇒at3+(2a−x1)t−y1=0→(1)
Now t1,t2,t3 are the roots of (1). ∴t1t2t3=y1a
Slope of the normal at t1 is −t1 , slope of normal at t2 is −t2
Normals at t1 and t2 are perpendicular ⇒(−t1)(−t2)=−1⇒t1t2t3=−t3⇒y1a=−t3⇒t3=y1a
t3 is a root of (1) ⇒a(−y1a)3+(2a−x1)(−y1a)−y1=0⇒−y31a2−(2a−x1)y1a−y1=0
⇒y21+a(2a−x1)+a2=0⇒y21=a(x1−3a) ∴ The locus of P is y2=a(x−3a)