Let P be (at2,2at) and Q be (h,k). Also given, A=(−2,3) Now, since Q divides AP in the ratio 3:2 i.e., ‌
AQ
QP
=‌
3
2
So, ‌‌h=‌
3at2−4
5
. . . (i) and k=‌
6at+6
5
. . . (ii) ⇒‌‌5k=6at+6 ⇒‌‌t=‌
5k−6
6a
put in Eq. (i), we get 5h=3a(‌
5k−6
6a
)2−4 ⇒‌‌5h+4=‌
(5k−6)2
12a
⇒(5k−6)2=60ah+48a ⇒‌‌(5k−6)2=12a(5h+4) Hence, the locus of Q(h,k) is (5y−6)2=12a(5x+4) . . . (iii) On comparing Eq. (iii) by Y2=4AX, we get, ⇒‌‌Y‌=5y−6,X=5x+4‌ and ‌4A=12a A‌=3a So, focus of Y2=4AX is (A,0) i.e. X=A and Y=0 Hence, focus of Eq. (iii) is 5x+4‌‌=3a‌ and ‌5y−6=0 x‌‌=‌