The correct option is A 343y2=12(x−6)3
Let P(at21,2at1) and Q(at22,2at2)
As it is given that 2×2at1=2at2
⇒t2=2t1
Now, point of intersection of normals at P and Q will be
(2a+a(t21+t22+t1t2),−at1t2(t1+t2))
But here a=3 and t2=2t1
Hence, point of intersection will be
(3(2+7t21),−18t31)≡(h,k)
t1=−3√k18
h=6+21(−3√k18)2⇒343k2=12(h−6)3
Hence, locus will be 343y2=12(x−6)3
Alternate Solution :
Let (h,k) is the point of intersection of the normals.
So, normal equation passing through (h,k) is
∴k+th=2at+at3 ...(i)
⇒at3+(2a−h)t−k=0
t1t2t3=ka, t1+t2+t3=0
Also given, 2at12at2=12
⇒2t1=t2⇒t3=−3t1
∴(t1)(2t1)(−3t1)=ka
⇒t31=−k6a
Since, t1 satisfies equation (i)
∴(2a−h)3t31=(k−at31)3
⇒−(2a−h)3k6a=(k+k6)3
∴−(6−x)3=34312y2
i.e., 343y2=−12(6−x)3=12(x−6)3