We have y2=12x ........(1) Comparing Eq. (i) with y2=4ax, we get 4a=12 ⇒a=3 ∴ Focus =(3,0) Let (x1,y1) be point on parabola Eq. (i), hence y12=12x1.....(ii) Now, according to question, √(x1−3)2+(y1−0)2=12 ⇒‌‌x12+9−6x1+y12=144....(iii) Putting y12=12x, from Eq. (ii) in Eq. (iii), we get x12−6x1+9+12x1=144 ⇒‌‌x12+6x1−135=0 ⇒‌‌(x1+15)(x1−9)=0 ⇒‌‌x1=9,−15 x1=−15 is not possible. From Eq. (i) y12=12x1 ⇒y12=12×9=108 ⇒y1=6√3 ∴ Required point =(9,6√3)