A(-2,3) is a fixed point outside the parabola y^2=4 a x(a>0) and P is a point moving on the parabola. The locus of point Q which divides A P in the ratio 3: 2 is a conic. Then focus of that conic is

Step-by-step solution

Let P be (a t^2, 2 a t ) and Q be (h, k) . Also given, A=(-2,3) Now, since Q divides A P in the ratio 3: 2 i.e., A Q Q P = 3 2 array rlrl & 5 k & =6 a t+6 t & = 5 k-6 6 a put in Eq. (i), we get 5 h & =3 a ( 5 k-6 6 a )^2-4 5 h+4 & = (5 k-6)^2 12 a (5 k-6)^2=60 a h+48 a (5 k-6)^2 & =12 a(5 h+4) array Hence, the locus of Q(h, k) is On comparing Eq. (iii) by Y^2=4 A X , we get, aligned & Y & =5 y-6, X=5 x+4 and 4 A=12 a & A & =3 a aligned So, focus of Y^2=4 A X is (A, 0) i.e. X=A and Y=0 Hence, focus of Eq. (iii) is a